Post on 25-Feb-2018
7/25/2019 Solucionario II
1/157
IIUN I DAD 1
OBRA COLECTIVA, DISEADA, CREADA Y PRODUCIDA
BAJO LA DIRECCIN DE:
ERLITA OJEDA ZAARTUDRA. EN CIENCIAS DE LA EDUCACIN
S o l u c i o n a r i o
S E C U N D A R I A
7/25/2019 Solucionario II
2/157
2 Matemtica II
Clave:a
Clave: e
Clave: c
Clave: c
Clave: a
Clave: b
Clave: d
Clave: d
Clave: c
Clave: e
Clave: c
1. II. La quiwicha es deliciosa y la maca tienegrandes propiedades
1. 800 1177 72,...3022 8
hay 72 nmeros mltiplos de 11
2. 200 1212 16,...8072 8hay 16 nmeros mltiplos de 12
2. p: 22+ 1 = 5 (V ) q: 32 1 = 6 (F)
(p q) (q p) (F F) (F V ) F V V
4. (p q) (r s) F
V F Luego: p F q F r V s V
5. (p q) (r s) F
V F
Luego: p V q F r V s F
6. (p q) (q r) V
~p q V ; q r V
p F
q Vr V
8. p: 23+ 32= 17 (V )q: 62= 36 (V )r: 32+ 42> 24 (V )
I. (p q) r
V V V
II. (p r) qV V
V
III. p (q r) V V V
9. (p q) (r p)
7. (p q) (r s) F
F F
Luego:p V r Fq V s F
3. p q
1U N I D A D
Pg. 10
Pg. 14
I. (p q) q F F F
II. (p q) (p s) V F
F
III. p (p r) V F F
7/25/2019 Solucionario II
3/157
Matemtica II
Clave: b
Clave: e
Clave: c
Clave: d
Clave: d
Clave: b
Clave: b
Clave: b
Clave: e
3. 187 < 6 < 794
187 < 6k < 794
31,1... < k < 132,3
k: 32, 33, 34, ...., 132
ktoma: 132 31 = 101 valoresLuego hay 101 nmeros
10. 120 = 233 5
S.D. = 23+1 12 1
31+1 13 1
51+1 15 1
S.D. = 15 4 6
S.D. = 360
11. 22n= 2n11n
C.D. = 25
(n + 1)(n + 1) = 25 n = 4
12. 3 + 1N 4 + 1 N = MCM(3; 4; 7) + 1
7 + 1 N = 84 + 1 N = 85
14. 4 a 7 2 5 8 = 11 + + +
4 + a 7 + 2 5 + 8 = 11
a 6 = 11
a = 6
15. x 2 y 3 y x = 45
Para que sea 5 x = 5 luego:
5 2 y 3 y 5 = 9
5 + 2 + y + 3 + y + 5 = 9
13. 4 a a 8 = 71 2 3 1
4 + 2a + 3a + 8 = 7
5a + 4 = 7
a = 2 a = 9
Luego "a" puede tomar 2 valores
4. 500 2346 21,... 40 23 17
hay 21 nmeros mltiplos de 23
5. Sean los nmeros impares (a 2); (a); (a + 2);luego:(a 2) + (a) + (a + 2) = 3a = 3
6. Sean los nmeros consecutivos (a 2); (a 1); a;(a + 1); (a + 2)
Luego:
a 2 + a 1 + a + a + 1 + a + 2 = 5a = 5
7. 12 = 223
hay 2 divisores primos
8. N = 243 55
C.D.(N) = (4 + 1)(1 + 1)(5 + 1)
= 60
9. 360 = 23325
C.D.compuestos= 24 3 1
= 20
Clave: dClave: c
Clave: c
5
9
7/25/2019 Solucionario II
4/157
4 Matemtica II
Clave: c
Clave: e
1. 50 175 5
10 35 5
2 7
MCD = 5 5 = 25
2. 16 48 96 15 2
8 24 48 15 2
4 12 24 15 2
2 6 12 15 2
1 3 6 15 2
1 3 3 15 3
1 1 1 5 5
1 1 1 1
3. MCM(7k; 3k) = 189
7 3 k = 189
k = 9
15 + 2y = 9 y = 6Piden:x + y = 11
Clave: dClave: a
Clave: e
Clave: aClave: c
Clave: c
Clave: d
16. C.D.(N) = 12
N = 2a 3b
C.D.(N) = (a+ 1)(b+ 1)
12 = (a+ 1)(b+ 1)
3 4 = (a+ 1)(b+ 1)
a= 2 b= 3
a= 3 b= 2
Por dato debe ser el menor posible nmero es:
N = 23 32
N = 72
Piden: 7 + 2 = 9
20. Sea n el nmero de veces que se debe mulplicar por 8 a 300, luego:
300 8n
= 223 5223n
= 23n+23 52
Por dato:
C.D. = 126
(3n + 3)(1 + 1)(2 + 1) = 126 n = 6
17. 7422222 = 8 + r
120 cifras
Luego:
222 = 8 + r
8 + 6 = 8 + r
r = 6
18. 10 + 20 + 30 + 40 + 50 + 5a = 37
150 + 5a = 37
7
a = 7
19. 4n
4n2
= 4n 1 1
16
= 4n1516
= 4n215
= 22n43 5
Por dato: C.D. = 28
(2n 3)(1 + 1)(1 + 1) = 28
2n 3 = 7
n = 5
Pg. 18
MCM = 25 3 5 = 48
7/25/2019 Solucionario II
5/157
Matemtica II
Clave: b
Clave: a
Clave: c
Clave: c
Clave: c
Clave: d
Clave: c
Clave: a
Clave: b
Clave: a
Clave: b
10. MCD(A; B) MCM(A; B) = A B
2 240 = 16 B
30 = B
11. 20 8 210 4 2
5 2 2
5 1 5
1 1
Piden: 200 : 40 = 5
12. MCD(180; 234)
180 234 2
90 117 3 30 39 3
10 13
Longitud del lado = 232 = 18
N lotes: = = 130
13. MCD(A; B) MCM(A; B) = A B
MCD(A; B) 350 = 1750
MCD(A; B) = 5
14. 2 6 1 2
1 3 1 3
1 1 1
N de ladrillos =6 6 6
= 36 2 3 1
15. MCD(A; B) = 4
4 3 2 2
A B 20 8 4
20 8 4 0
28 n lapiceros
4. MCD (135; 90; 225)
135 90 225 5
27 18 45 3
9 6 15 3
3 2 5Piden:
3 + 2 + 5 = 10
5. MCD (140; 168; 224)
140 168 224 4
35 42 56 7 5 6 8
n cajas: 5 + 6 + 8 = 19
6. MCD(12k; 18k; 30k) = 48
6k = 48
k = 8
7. A = 243654113
B = 23385272
MCM(A; B) = 24 385472113
8. A = 25337
B= 337 11
MCD(A; B) = 337
9. 40 + 3
N 30 + 3
16 + 3
N = 240 + 3
Nmnimo= 243
N = MCM(40, 30, 16) + 3
MCM = 23 5 = 40
6
AtotalAlote
180 234182
7/25/2019 Solucionario II
6/157
6 Matemtica II
Clave: c
Clave: e
Clave: e
Clave: b
Clave: e
Clave: cClave: b
Clave: b
Clave: a
6. 20 4 = 4 c
12 = c
324
= 40d
d = 5
Piden: 12 5 = 17
7.AB
= 29
A = 2k, B = 9k
Por dato:
9k 2k = 84
k = 12
Piden: B = 9(12) = 108
8.a8
= b10
= c18
= k
a = 8k, b = 10k, c = 18k
Por dato:
a + b + c = 180
8k + 10k + 18k = 180
k = 5
Piden: a = 8k
a = 40
9.AB
= 58
A = 5k, B = 8k
Por dato:
(5k)2+ (8k)2= 356
25k2+ 64k2= 356
89k
2
= 356 k = 2
Piden: A B = (5k)(8k)
= 40k2
= 160
B = 20 3+ 8
B = 68
A = 68 4 + 20A = 292
Clave: d
1. Total: 400 personas
Mujeres: 240
Varones: 160
VM
=160240
= 23
2.A = 3k, B = 4kPor dato:7k = 56k = 8
Piden: B = 4k = 32
3. A = 3k, B = 2k
Por dato:
3k 2k = 5
k = 5Piden : B = 2k = 10
4. Pedro=
180=
9
Jos 80 4
5. a = 3k
b = 7k
Por dato:
a + b = 80
10k = 80
k = 8
Piden: b = 7k = 56
Pg. 22
7/25/2019 Solucionario II
7/157
Matemtica II
Clave: a
Clave: a
Clave: c
Clave: e
10.Total: 400 personas
HM
= 3k2k
= 21
3k + 2k = 400
k = 80
V = 240; M = 160
Luego:240 x160 x
= 21
x = 80Luego, se retiraron 80 parejas
11. AlcoholAgua = 53 Alcohol = 5k; Agua = 3k
Por dato:
8k = 72
k = 9 Alcohol = 45
Luego:
4527 + x
= 910
x = 23
12.AB
= 811
= 79
8k + 1011k + 10
= 79
72k + 90 = 77k + 70
20 = 5k
4 = kPiden:
8k 411k 4
= 32 444 4
= 2840
= 710
Luego, la relacin era de 7 a 10
13.AB
= 5k8k
= 79
5k + 228k + 22
= 79
45k + 198 = 56k + 154
4 = k
A = 5k = 20 , B = 8k = 32
15.AB
= 7k5k
= 1
Luego:
7k 4 = 5k + 4
2k = 8
k = 4
A = 7k = 28; B = 5k = 20Piden:
A + B = 48
14.LA
= 95
L = 9k; A = 5k
Por dato:
2L + 2A = 336
18k + 10k = 336
28k = 336
k = 12
L A = (9k)(5k)
= 45(12)2
= 6 480
Clave: c
Clave: b
x+22
4
+22
+4
x
+ 10 aos
+ 10 aos
7/25/2019 Solucionario II
8/157
8 Matemtica II
Clave: cClave: c
Clave: aClave: c
Clave: c
Clave: c
Clave: c
Clave: b
Clave: a
Clave: d
Clave: d
Clave: d
1.15
100 60
1001 200 = 108
2.20
100
30100
60
1009 000 = 324
3.80
100
75100
100100
30 = 18
4. Tengo 100%
Vendo 20%
80% Vendo 40%
40%
Luego: 40% N = 200
N = 500
8. Au = 10 + 20 + 10 20100
%
Au = 32%
9. 100% 15% = 85%Sea N el precio de la grabadora, luego:
85% N = 170
N = 200
10.Tengo: 200
20% (200) = 40
Luego tengo:
200 + 40 = 240
Gast: 20% (240) = 48
Queda:
240 48 = 192
Piden: 200 192 = 8
11. Gano: 40% (400) = 160
Tengo:560
Luegogasto: 20% (560) = 112
Queda: 560 112 = 448
Al final gano (S/.)448 400 = 48
12. Nios: 40%
Nias: 60%
Ahora:
Nios: 20%
Nias: 60%
80%
80% 100%
60% x%
x = 100 6080
x = 75%
5. 200 100%
120 x%
buen estado
x = 120
100200
x = 60%
7. Du = 20 + 40 20 40100
%Du = 52%
6. 115 son papayas
345 no son papayas
460 100%
345 x%
x = 345 100460
x = 75%
Pg. 28
460
7/25/2019 Solucionario II
9/157
Matemtica II
1. V = 7k; M = 9k
Por dato:
16k = 160
k = 10
Luego: V = 70; M = 90
Entonces deben llegar 20 varones
2. N < 600
10 + 7
N 12 + 7 N = MCM(10; 12; 15) + 7 15 + 7
N = 60 + 7
N = 540 + 7
N = 547
Clave: c
Clave: b
Clave: c
Clave: a
6. V1= 10 ; D1= 15
V2= 5 ; D2= x
Se cumple:
V1D1= V2D210 15 = 5 D2
30 = D2 Clave: a
Clave: e
Clave: e
13. 540 15% = 81
Me deben S/. 81
Pagan: 20% 300 = 60
An me deben: 81 60 = 21
14. 210 100%
x 30%
x = 210 30100
x = 63
Piden:
210 + 63 = 273
15.
Solucin de problemas Pg. 30
3. obreros h/d das zanja 3 4 6 12 x 6 2 16
3 4 612
= x 6 216
x = 84. n secretarias digitan da
2 350 7
x 600 42 7350
= x 4600
x = 6 Clave: d
5. Precio D.P. (tamao)2 I.P. Energa
P ET2
= k
360 M(14)2
=P
(21)2
360 M196
=
P
441
360 M 441 2
196 M= P
1 620 = P Clave: d
M4
M4
7. Lado = 3k ; Lado menor = 2k
A = A
(3k)2
= (2k) x 9k2= 2k x
4,5 k = x
PermetroPermetro
= 4(3k)2(2k) + 2(4,5k)
= 12k13k
= 1213
Clave: a
A1= abA2= (60%a)(140%b)A2= 84%abA2= 84% A1Luego, el rea disminuye en 16%
A1 A2 60%aa
b 140%b
7/25/2019 Solucionario II
10/157
10 Matemtica II
13. A = 47 + 5
B = 47 + 31
C = 47 + 41
A + B + C = 47 + 77 = 47 + (47 + 30)
= 47 + 30
residuo = 30
14. Pc = S/. 600
Pv = Pc 25% Pv
125% Pv = 600
Pv = 480
Pg. 33
10. Sea la cantidad de dinero "x", por dato:
Se pierde: 40% x queda 60% x
Se gana: 20%(60%x)
Se tiene: 120%(60%x)
Luego:
120%(60%x) 88 = 200
72x100
= 288
x = 400
8. 100% + 60% = 160%
3N 160% 100%
N 1603
% x
x =
1603
100160
x = 33,3%
9. El equipo gan 12 encuentros, luego:
12 + x 100%
x 85%
x = (12 + x) 85100
x = 68
1. (p q) r F
p V
q V
r F
2. I. V F V
II. F F V
III. V F F
IV. F F F
3. 999 = 9 111
= 323 37
= 3337
C.D.(999) = (3 + 1)(1 + 1)
= 8
Clave: a
Clave: a
Clave: c Clave: c
Clave: e
Clave: e
Clave: e
Clave: d
Clave: b
Clave: b
11. 36 = 720 30 t1 200
t = 2
12. Pv = 6 500
Pv = Pc + G
6 500 = 130% Pc
5 000 = Pc
7/25/2019 Solucionario II
11/157
Matemtica II
10. A B = 245
MCM(A; B) = 5MCD(A; B)
Sabemos que:
MCD(A; B) MCM(A; B) = A B
MCD(A; B) 5MCD(A; B) = 245MCD(A; B) = 7
MCM (A; B) = 35
A = 35
B = 7
Piden: 35 7 = 28
11.HP
= 35
= 57
3k + 205k + 20
= 57
21k + 140 = 25k + 100
40 = 4k
10 = k
Piden: H = 3k = 30
12. a24
= 24c
= 23
a
24= 2
3 a = 16
24c
= 23
36 = c
Piden:a + c = 52
13. rea das
72 8
142 x
x = 142 872
x = 32
4. 240 = 243 5
C.D.(240) = 5 2 2
= 20
C.D.(N) = C.D.S+ C.D.C
20 = 4 + CDC 16 = C.D.C
5. 9 999 = 9 11 101
= 3211 101
Piden:
3 + 11 + 101 + 1 = 116
6. 1 800 = 233252
C.D.(1 800) = 4(3)(3)
= 36
7. 820 = 225 41
= 4 (5 41)
C.D.(4) = (2)(2)
= 4
8. N = 412 410
N = 410 (42 1)
N = 220(16 1)
N = 2203 5
C.D.(N) = (21)(2)(2)
C.D.(N) = 84
9. AB = 6 A = 6B
MCM(A; B) = 72
MCM(6B; B) = 72
6B = 72
B = 12
Clave: c
Clave: c
Clave: c
Clave: c
Clave: cClave: b
Clave: e
Clave: e
Clave: a
Clave: d
+ 20 aos
+ 20 aos
Luego:
7/25/2019 Solucionario II
12/157
12 Matemtica II
1. A = 5k, B = 7k, C = 11k
Luego de sumar 130, 260 y n a cada uno deellos, respectivamente tenemos que:
(*)
5k + 130=
7k + 260=
11k + n
13 17 19
k = 195Reemplazamos en (*)
7(195) + 26017
= 11(195) + n19
Resolviendo n = 910
3. Sea : divisor comn de 2 520 y 2 000 y el myor posible (para usar menos trabajadores)
= MCD (2 520; 2 000) = 40 m
2 520 m
2 000 m
N murales
1 av: 2 52040
+ 1 = 63 + 1 = 64
2 av: 2 00040 + 1 = 50 + 1 = 51 115 murales
Como se necesitan 3 trabajadores por mural
115 3 = 345
2. Inicialmente: 8 personas / 10 das / 8 h/d
obra
Se hizo as: 8 personas (8 + x) personas 5 das 2 das
5 h/d 10 h/d
Se observa que las 8 personas han trabajado mitad del tiempo indicado para concluir la obpor lo tanto, solo han hecho la mitad del trabjo. En consecuencia ahora todos los obreros coel grupo que se incorpora debern terminar obra, es decir, debern realizar la mitad del trbajo.
Adems, recordemos que:
(N de obreros)(N de das)(N de h/d) = cte.
Reemplazando valores, tenemos que:
8 5 8 = (8 + x) 2 10
x = 8
14. Peones h/d das obra (m2)
14 7 15 150
21 8 x 240
14 7 15150
= 21 8 x240
x = 14
15. Pv = 2 600
Pv = Pc + G
Pv = 100% Pc + 30% Pc
2 600 = 130% Pc
2 000 = Pc
16. C = 4 000
r = 7,5% trimestral 30% anual
t = 15 meses
I= 4000 30 151 200
I= 1 500
Clave: b
Clave: a
Clave: aClave: c
Clave: c
Clave: c
Se contratarn x personasadicionales
7/25/2019 Solucionario II
13/157
Matemtica II
4. a + b + c = 12 abc = 12
bc = 4 (criterio por 4)
Como c es par:
c = 0 b = 4; 6; 8
c = 2 b = 1; 3; 5; 7; 9
c = 4 b = 2; 4; 6
c = 6 b = 1; 3; 5
c = 8 b = 0; 2
Luego:
3 + 5 + 4 + 3 + 2
= 17 soluciones
7/25/2019 Solucionario II
14/157
14 Matemtica II
10.ab
ba
= 1,805
a2 b2
ab=
1805 180900
a2 b2
ab= 65
36
ab = 9 4
a + b = 13
11. Gasta = 34
no gasta
G N
G + N = 49
34
N + N = 49
74
N = 49
Clave: a
Clave: d
Clave: d
Clave: b
Clave: bClave: b
Clave: b
Clave: d
Clave: d
Clave: a
1. Sea R el recorrido y F lo que falta recorrer, pordato:
15
R =35
F R + F = 12 ... (I)
R = 3FRemplazamos en (I) 4F = 12
F = 3 R = 9
2. Sea x el nmero de aves:
Palomas:4x5
Otros:x5
Gallinas:x5
56
Gallos16
x5
= 8
x = 2403. Sea x lo que me deben, luego:
x =78
960 = 840
Ahora:
Pagan:34
(840) = 630
Me deben: 840 630 = 210
4.12
100+
39
+582 58
900
108 + 300 + 524900 = 932900 = 233225
5. E =997 99
9+
29
E =898 + 2
9=
9009
= 100
Piden: 100 = 10
6. 23
910
56
+ x = 59
1 + 2x
2=
59
9 + 18x = 10
x =1
18
7. x =4 5
20= 1
20
y =5 4
20=
120
8. x +7
15= 10
15
x = 315
x =15
9. Tipo A Tipo B
Usan Queda Usan Queda
56
16
34
14
16
+ 14
=2 + 3
12= 5
12
2U N I D A D
Pg. 40
y > x
7/25/2019 Solucionario II
15/157
Matemtica II
12. 35
3= 27
125
1625
25
1125
=80 50 1
125=
29125
27125
+29
125=
56125
1
56
125 =
69
125
13.7098
= 5k7k
(5k)(7k) = 315
k2= 9
k = 3
Luego: 5(3)
7(3)
= 15
2121 15 = 6
15. M = 1 1
1 1 +
118
M = 1 1 +
187
M = 1 7
15=
815
16. E =3 + 1
12
1312
6 + 4 + 312
7413
E =
13123712
7413
E =1337
7413
= 2
17. P =1
+
2 +
159 73
90119
=1
+
2 +
159 73
110
P =1
+59 110293
=1
1465 + 9902637
P =1
24552637
=26372455
1,074
14. M = a + b
11a + 7b77
=844 155999 999
11a + 7b =77 844 155
999 999
11a + 7b = 65
4 3
a + b = 7
Clave: e
Clave: e
Clave: b
Clave: a
Clave: d
Clave: d
Clave: c
N = 28
G = 34
(28) = 2178
157
7/25/2019 Solucionario II
16/157
16 Matemtica II
18. E =321 32
900249 24
900
1
E = 289900
225900
1
E =1730
1530
1
=2
30
1
= 15
19. - Un litro de leche cuesta S/.17,50
- Un litro de vino cuesta S/.43,75
43,75 : 17,50 = 2,5
26.
Total: 5050
25
= 20
50 20 = 30
30 + 15 = 45
4523
= 30
Ahora:
45 30 = 15
15 + 35 = 50
20. 8431,24 : 0,97 = 8 692
21. 503,54 84,5 = 419,04
419,04 : 0,97 = 432
22. 846,40 : 32 = 26,45
El metro cuesta 26,45
Piden 20 m
26,45 20 = 529
23. 750 : 2 400 = 0,3125
0,3125 84 = 26,25
24. 2,60 12 = 31,2
31,2 : 13 = 2,4
25.x3
+ 40 = + 40x 2x3
x3
x3
+ 40 + 40 = 8425
Clave: e
Clave: d
Clave: a
Clave: b
Clave: c
Clave: c
Clave: c
Clave: b
Clave: b
Clave: d
Clave: c
3x + 360 = 1 260
x = 300
1. FFVF
3.
4.
2.12
= 0,5
34
= 0,75
3 = 1,73 p= 3,14
2 2 = 2,82
Ordenamos de mayor a menor
p; 2 2 ;12
; 34
; 3
Pg. 44
2,8 + 0 10 3 + 1523
p
2
B A = 3; +
+ 32
AB
7/25/2019 Solucionario II
17/157
Matemtica II
Clave: a
Clave: e
Clave: b
Clave: a
Clave: b
Clave: c
Clave: c
Clave: a
Clave: b
Clave: d
Clave:d
Clave: d
10. |x + 1| = 0
x + 1 = 0
x = 1
11. |3x 2| = 0
3x 2 = 0
x =23
12. |x 3| = 2
x 3 = 2 x 3 = 2
x = 5 x = 1
C.S. = {1; 5}
13. II y III
14.
15.
16.
17.
5.
6.
7.
8.
9.
A B = [1; 5]
A B = [1; 37; 9
B A = [2; 1
Q P = ; 5
M N = [8; 2]
M N = [4; +
A B =
A B = [2,3; 3,1
+
+
+
+
+
+
+
+
2
7 9
2
9
8
5
0
3
1
5
2
2
8
4
3
3,1
1 5
1
2
0
5
6
2
2,3
A
A
A
Q
M
M
A
A
B
B
B
P
N
N
B
B
1 + 2 3
2 p 3,777
4 5
e + 1 5 + 2
158
7/25/2019 Solucionario II
18/157
18 Matemtica II
Clave: b
Clave:
c
Clave: a
Clave: a Clave: b
Clave: b
Clave: e
Clave: e
18. 5 2x 1 < 4
+1 4 2x < 5
: 2 2 x 0)
y =7
x(x 4)
x 0 x 4 0
x 4
Dom = {0; 4} ...............(V )
Rpta: V F V
9. 2a + 1 = 5
2a = 6 a = 3
b = b + 4
2b = 4 b = 2
Piden:
3a 2b = 3(3) 2(2) = 9 + 4 = 5
8. g(x) = 7 + (3x 6)
3x 6 0
3x 6
x 2
Evaluamos para x = 2
7 + (3(2) 6))
7 + 0 = 7
Luego, Rango = [7, +
2
+ +
0 1
12
12
7/25/2019 Solucionario II
33/157
Matemtica II
12. f: funcin cuadrtica
f(x) = ax2+ bx + c
f(1) = 2
f(1) = 2f(2) = 4
Reemplazando:
a + b + c = 2 ...(I)
a b + c = 2 ...(II)
4a + 2b + c = 4 ...(III)
De (I) y (II)
2a + 2c = 0
a + c = 0 a = c
De (II) y (III)
2a 2b + 2c = 4
4a + 2b + c = 4
6a + 3c = 8
6(c) + 3c = 8
6c + 3c = 8
3c = 8
c =83
a = 83
En (I):
a + b + c = 2
83
+ b +83
= 2
b = 2
Reemplazamos en la funcin:
f(x) = 83
x2+ 2x + 8
Clave: b
Clave: a
Clave: a
Clave: a
13.f(x) = 42x 1 + 6 3x
Analizamos por partes:42x 1 2x 1 0
x 1
2 6 3x 6 3x 0
3x 6
3x 6
x 2
0 1/2 1 2
x [1/2; 2]
14.Ran(f ) = {3; 22; 59} , f(x) = x3 5
x3 5 = 3 x3 5 = 22 x3 5 = 5
x1= 2 x2= 3 x3= 4
Dom(f) = {2; 3; 4}
Piden: 2 + 3 + 4 = 9
15.f = {(2; 3), (3; 6), (2; 3a + b), (3; b)}
Se cumple:
3 = 3a + b b = 6
3 = 3a + 6
3 = 3a
a = 1Piden:
a + b = 1 + 6
= 5
7/25/2019 Solucionario II
34/157
34 Matemtica II
Clave: c
Clave: d
16.f(x) =1
x2 x
x2 x 0
x(x 1) 0
x 0 x 1 0 x 1
Dom f = {0, 1}
17. I. y = 2x + 7 (es funcin)
II. y = x2 2 (es funcin)
III. x2+ y2= 16 (no es funcin)
Rpta: Iy II
20. f(x) = x2+ x 2
f(3) = (3)2+ 3 2 = 10
f(4) = (4)2+ 4 2 = 18
f(5) = (5)2+ 5 2 = 28
f(2) = (2)2+ 2 2 = 4
f(1) = (1)2 + 1 2 = 0
f(0) = (0)2+ 0 2 = 2
Verificando:
I. f(3) + f(4) f(5) = 0
10 + 18 28 = 0 ... (V)
II. 2f(2) f(0) = f(3)
2(4) (2) = 10
8 + 2 = 10 ... (V )
III. 5f(2) 2f(3) = f(1)
5(4) 2(10) = 0
20 20 = 0 ... (V )
Rpta: VVV
18.x2 2x + 3 =x2
+ 2
2x2 4x + 6 = x + 4
2x2 5x + 2 = 0
x1= 2 x2 =12
Si x1= 2 y =22
+ 2 = 3P(2; 3)
Si x2=
1
2
y =
1
4 + 2 =
9
4
Q
1
2 ;
9
4
Suma de coordenadas:
2 + 3 +12
+94
=
8 + 12 + 2 + 94
=314
21.
Rpta: Forma un rectngulo
19.
Clave: c
Clave: b
Clave: e
Clave: d
Dom (f) = [2;3]
Ran (f) = [4; 5]
Dom (f) Ran (f) = [2; 3]
5
10
y
y
x
x
f
3
6
4
0
2
y = 10 x = 6f(y) = 6
f(x) = 10
7/25/2019 Solucionario II
35/157
Matemtica II
22. f(x) = mx + b ... (I)
7 = f(4) = 4m + b
1 = f(3) = 3m + b
7 = 4m + b
1 = 3m b ()
6 = m
Reemplazamos en (I)
f(x) = 6x 17
Piden: f(2) = 6(2) 17
f(2) = 29
Piden Dom(f) Dom(g)
= [3; + {3}
= 3; +
A = b h2
A =10 5
2
A = 25
24. f(x) = x 3
x 3 0
x 3
Dom(f) = [3; +
g(x)=7 4x
x 3
x 3 0
x 3
Dom(g) = {3}
Clave: a
Clave: d
Clave: b
b = 17
5
y x = 5
(5; 5)
(5; 5)
x5
5
0
23.Sea : {(x, y} 2/x = |y| } y x = 5
Clave: d
Clave: c
1. Iy IV
2. 8x3y + 6x3y 10yx3= 12x3y
Pg. 77
Clave: a
Clave: a
Clave: c
Clave: a
3. {[7a2b3+ 8a3b2] 5a2b3} + 8a3b2
= {+7a2b3 8a3b2 5a2b3} + 8a3b2
= 2a2b3 8a3b2+ 8a3b2
= 2a2b3
4. P = 3x2 x4 + (2x2)3 100x12
P = 3x6+ 8x6 10x6
P = x6
Piden:
P2= (x6)2= x12
5. (3x6y5)2 81x24y20
9x12y10 9x12y10= 0
6.tpq10
33;
83
pt 10qx
Por trminos semejantes se cumple:
1 = t 10 t = 11x = 10
Luego los coeficientes son:1133
y83
Piden:1133
+83
= 3
7/25/2019 Solucionario II
36/157
36 Matemtica II
Clave: e
Clave: b
Clave:
b
Clave: e
Clave: d
10. P(x) = 3x + 5
P(P(x) = 3(P(x) 5) + x
P(3x + 5) = 3(3x + 5 5) + x
3(3x + 5) + 5 = 3(3x) + x
9x + 15 + 5 = 9x + x
20 = x
11. A =G.A. (P) + G.R. (z)
G.R. (x) G.R. (y)
A =12 + 5
4 3
A = 17
9.
P(1; 2) = 4(1)(2)
3
6(1)
2
(2) = 32 + 12 = 44
7. 9x5+ (30x13 : 5x10)2+ 4 28x10
7x83
+ x4
9x5+ (6x3)2+ 4 (4x2)3 + x4
9x5+ 36x6+ 4 64x6+ x4
Ordenando:28x6+ 9x5+ x4+ 4
8. Si P(x 2) = 3x + 4
x 2 = 5
x = 7
P(5) = 3(7) + 4 = 21 + 4 = 25
x 2 = 3
x = 5
P(3) = 3(5) + 4 = 15 + 4 = 19
x 2 = 4
x = 6
P(4) = 3(6) + 4 = 189 + 4 = 22
Piden:
P(5) + P(3) P(4)
= 25 + 19 22 = 22
Clave: c
Clave: e
Clave: b
Clave: c
Clave: d
12. 3x7yz6 + a
Por dato, G.A. = 16
7 + 1 + 6 + a = 16
Piden: a = 2
(2)3= 8
13. x6a + 4y3a + 6z4a + 2
Por dato, G.A. = 38
6a + 4 + 3a + 6 + 4a + 2 = 38
13a + 12 = 38
a = 2
Luego el monomio es: x16y12z10
G.R.(y) = 12
14. G.A. = 14 a + 6 = 14 a = 8
Piden:
a2 3a + 1 = 82 3(8) + 1 = 41
15. Por dato:
G.A. (M) = 23
2a + 3a 1 + 4a 2= 23
3 9a 3 = 69
a = 8
Piden: [2(8)]2= 256
16. Por dato:
G.R(x) = 14
4a + b = 14 ... (I)
G.A. (M) = 18
14 + 2a b = 18
2a b = 4 ... (II)
Sumamos (I) y (II)
6a = 18
a = 3 b = 2
Piden: a + b = 5
7/25/2019 Solucionario II
37/157
Matemtica II
Clave: cClave: d
Clave: b
Clave: a
Clave: b
Clave: e
Clave: c
Clave: c
17. Por dato:
P(2) = 7
2a + b = 7 ... (I)
P(3) = 8
3a + b = 8 ... (II) Restamos (I) y (II)
5a = 15
a = 3
21. P(x) = 5ax2+3bx 6 10x2+ 9x + 2c
= (5a 10)x2+ (3b + 9)x + (2c 6)
Como P(x)0 , se cumple:
5a 10 = 0 a = 2
3b + 9 = 0
b = 3 2c 6 = 0 c = 3
Piden :
b2+ c2=
(3)2+ 32= 9
a 2
22. Dato:
G.A.(P) G.R.(x) + 2G.R.(y) = 24
2n + 3 ( n + 4) + 2( n + 2) = 24
3n + 3 = 24
n = 7
Piden : G.R.(x) + G.R.(y) = n + 4 + n + 2= 2n + 6
= 2(7) + 6
= 20
23. R = P(40) P(P(P(2)))
R = 3(40) + 5 P(P(11))
R = 125 P(38)
R= 125 119
R = 6
24. P(x) = 4x3 6ax + (2b 5)
Por dato: T.I. = 7
2b 5 = 7
b = 6
Luego, P(x) = 4x3 6ax + 7
Por dato: P(2) P(1) = 18
(32 12a + 7) (4 + 6a + 7 ) = 18
39 12a 3 6a = 18 a = 3
Piden : ( 5b a)1/a= (30 3)1/3
= 3
18. Por dato: P(Q(Q(2))) = 1 ... (I)
Calculamos: Q(2) = 2 + 3
= 5
Reemplazamos en (I)
P(Q(5)) = 1
Calculamos Q(5) = 5 + 3 = 8
Reemplazamos: P(8) = 1
82 5(8) + m = 1
m = 23
19. 13x 15 = 2ax + 5a + bx 7b
13x 15 = (2a + b)x + (5a 7b)
2a + b = 13
(x7) 14a + 7b = 91 ...(I) 5a 7b = 15 ...(II)
Sumamos (I) y (II)
19a = 76
a = 4 b = 5
Piden: a b = 20
20. P(x) = 4x + 2x4+ 6mxm-5 3x3 4
Como el polinomio es completo, entonces:
m 5 = 2
m = 7
Piden:
4 + 2 + 6m 3 4
= 6m 1
= 6(7) 1
= 41
7/25/2019 Solucionario II
38/157
38 Matemtica II
Clave: a
1. {m + [( 4m + 8) 5m] + 9} + 8m
{m + [ 4m + 8 5m] + 9} + 8m
{m + [ 9m + 8] + 9} + 8m
{m 9m + 8 + 9} + 8m
{8m + 17} + 8m
8m + 17 + 8m = 17
Pg. 82
Clave: c
Clave: b
Clave: e
Clave: d
25. Por P. homogneo, se cumple:
a + 2 + 1 = 4 + b 2
a b = 1
Por dato:
b
2
a
2
= 5 (b a) (b + a) = 5
1
b + a = 5
a = 2 , b = 3
Luego: P(x; y) = 2x4y + 3x4y
= 5x4y
P(a; b) = P(2; 3)
= 5(2)4(3)
= 290
Piden: 290 = 5
48
28. Q(x) = m(x2 3x + 2) + n(x2 5x + 6)
+ p(x2 4x + 3)
Q(x) = (m + n + p)x2 (3m + 5n + 4p)x
+ (2m + 6n + 3p) Por dato: P(x) Q(x)
m + n + p = 7
3m + 5n + 4p = 6
2m + 6n + 3p = 1
Resolviendo:
m = 41 ; n = 7 ; p = 41
Piden:
m + n p = 41 + 7 (41)
= 89
26. Por P. homogneo, se cumple:
a2 + 3 + 2a = 3a +5
a2 a 2 = 0
a 2
a 1
a = 2 a = 1 ( no cumple)
Piden: (3a 1) (5a a3)
= a3 2a 1
= (2)3 2(2) 1
= 3
27. 8x 3 = 2ax a + bx + 3b
8x 3 = (2a + b)x + (3b a)
2a + b = 8 ... (I)
3b a = 3 (x2) 6b 2a = 6 ... (II)
Sumamos (I) y (II)
7b = 2 b =2
7
Reemplazamos en (I)
2a + 2
= 8 7
a =
27
7Piden : a + b =
27+
2=
29
7 7 7
7/25/2019 Solucionario II
39/157
Matemtica II
10. (5x2 6x 7) (x2+ 2x 1)
= 5x4+ 10x3 5x2+ 6x3 12x2+ 6x + 7x2
14x + 7
= 5x4+ 16x3 10x2 8x + 7
Piden:
5 + 16 10 8 + 7 = 0
2. A(x) = 4x2+ 3x 2
B(x) = 3x2+ 8x + 9
3A(x) 2B(x) 18x2+ 9x + 24
3(4x2+ 3x 2) 2(3x2+ 8x + 9) 18x2+ 9x + 24
12x2+ 9x 6 + 6x2 16x 18 18x2+ 9x + 24
18x2 7x 24 18x2+ 9x + 24 = 2x
3. 5x4 3x8 + 36x16 : 4x4 (2x3)4 8x12
15x12+ 9x12 16x12 8x12 = 0
4. [(2x3)4 : 8x10+14x5
7x3 3x2]8
[16x12 : 8x10+ 2x2 3x2]8
[2x2+ 2x2 3x2]8
[x2]8= x16
8. A = 8x 3x5+ 84x9 : 2x3 25x6 (3x2)3+ 2x6
A = 24x6+ 42x6 25x6 27x6+ 2x6
A = 16x6
Piden:
(A : 4x5)3= (16x6 : 4x5)3= (4x)3= 64x3
9. 2[P(x) + Q(x)] = 2[3x2+ x + 3)
= 6x2+ 2x + 66.
83
x6y5 72
x2y638 x
3y4
72
x6 + 2 3y5 6 + 4 = 72
x5y3
Reemplazando:
a = 7
2 , m = 5 , n = 3
Piden:
72
+ 15
13
= 105 + 6 1030
= 10130
7. A(x) = 2,3 + 0,4x2 + 3,8x
B(x) = 359
x + 2310
+ 25
x2
Usando generatriz:
A(x) = 25
x2 + 359
x + 2310
B(x) = 25
x2+ 359
x + 2310
Piden:
E = 4[A(x) B(x)]
E = 4 25
x2+ 3510
x + 2310
25
x2 3510
x 2310
E = 4[0] = 0
5.42xn + 8ym + 5z4 + p
6x6 + ny2 mz3 + p
= 7xn + 8 6 nym + 5 2 + mz4 + p 3 p
= 7x2y3z
Clave: c
Clave: c
Clave: b
Clave: b
Clave: e
Clave: e
Clave: e
Clave: a
Clave: d
7/25/2019 Solucionario II
40/157
40 Matemtica II
13.A = (5x + 2) (2x + 3) = 242
10x2+ 19x + 6 = 242
10x2+ 19x 236 = 0
10x + 59
x 4
(10x + 59) (x 4) = 0
x = 4
largo () = 5x + 2
= 5(4) + 2
= 22
ancho (a) = 2x + 3
= 2(4) + 3
= 11
Permetro = 2+ 2a = [2(22) + 2(11)]
= 66
17.(3x 4)(2x + 1) (3x + 2)(x 6) 3x(x 5)
6x2+ 3x 8x 4 (3x2 18x + 2x 12) = 3x2 + 1
6x
2
+ 3x 8x 4 3x
2
+ 18x 2x + 12 3x
2
+ 15x26x + 8
Clave: d
Clave: e
11. V = (7x + 3) (6x + 4) (5x + 2)
V = (42x2+ 46x + 12) (5x + 2)
V = 210x3 + 314x2 + 152x + 24
Para x = 1
V = 210 + 314 + 152 + 24
V = 700
12. A = (2x + y)(x y)
A = 2x2 2xy + xy y2
A = 2x2 xy y2
14. M(x) S(x) = D(x)
S(x) = M(x) D(x)
= 2x2+ 6x 3 (4x + 7)
= 2x2+ 2x 10
Clave: a
Clave: d
Clave: d
15. LO= 2(x 1)
= (2x 2)
16. A(x) B(x) = (2x + 5)(x 8)
ax2+ bx + c = 2x2 11x 40
a = 2 , b = 11 , c = 40
Piden : a + b c = 2 11 (40)
= 31
19. (3x + 2)(x 6) (2x 1 )(x 2)
= 3x2 18x + 2x 12 2x2+ 4x + x 2
= x2 11x 14
Piden: menor coeficiente = 14
20. Por dato:
G.R.(x) = 7
a + 5 = 7
a = 2
G.R.(y) = 6
b + 3 = 6
b = 3
18. A =3x2+ 2y2+ x2 y2
4x2y 2
A = (4x2+ y2)2x2y
A = 8x4y + 2x2y3
Clave: c
Clave: d
Clave: c
Clave: e
7/25/2019 Solucionario II
41/157
Matemtica II
22.R(x; y) = 8x2y4(5x4y2 6x3y4 x5y)
3x3y3(2x5y3+ 3xy6 2x4y5)
P(x; y) = 40x6y6 48x5y8 8x7y5+ 6x8y6
9x4y9+ 6x7y8
Luego G.A.(R) = 15
23.
A = [2(4x + 3y)] [3(5x 2y)]
A = (8x + 6y)(15x 6y)
A = 120x2 48xy + 90xy 36y2
A = 120x2+ 42xy 36y2
24.
V = (2x + 1)(x + 4) (3x 2)
V = (2x2 + 8x + x+ 4)(3x 2)
V = (2x2+ 9x + 4)(3x 2)
V = 6x3 4x2+ 27x2 18x + 12x 8
V = 6x3+ 23x2 6x 8
21.P(x) = 4x 3
Q(x) = 7x2 6x
R(x) = x3+ 5
5x3P(x) + 3x4R(x) + 2Q(x)
5x3 (4x 3) + 3x4(x3+ 5) + 2(7x2 6x)
20x4 15x3+ 3x7+ 15x4+ 14x2 12x
3x7+ 35x4 15x3+ 14x2 12x
G.R.(x) = 7
Clave: d
Clave: d
Clave: e
Clave: c
5x 2y
4x + 3y 2(4x + 3y)
3x 2
2x + 1x + 4
3(5x 2
Clave: b
Luego:
P(x; y) = x7y3+ 2x6y6+ 3x5y4
P(b; a) = P(3; 2)
= (37)(23) + 2(36)(26) + 3(35)(24)
= 17 496 + 93 312 + 11 664
= 122 472
P(a; b) = P(2; 3)
=(27)(33) + 2(26)(36) + 3(25)(34)
= 3 456 + 93 312 + 7 776
= 104 544
Piden:
122 472 104 544 429
= 17 499
Clave: b
Solucin de problemas
1. A R B
H. Valdizn Cercado de Lima
A. Loayza Miraflores
J.C. Ulloa AteVitarte
R = {(Hermilio Valdizn, Ate Vitarte),
(Arzobispo Loayza, Cercado de Lima),
(Jos Casimiro Ulloa, Miraflores)}
Pg. 84
7/25/2019 Solucionario II
42/157
42 Matemtica II
Clave: a
Clave: b
Clave: e
Clave: d
Clave: d
Clave: a
2. Distancia: d
Tiempo: t
d = 80t
Para t = 4,
d = 80(4)
d = 320
3. T: tarifa (S/.)
M: Masa (g)
Arequipa: T1= 3,50 + 0,75 m
Cusco: T2= 5,00 + 1,25 m
Trujillo: T3= 2.50 + 0,40 m
T1 = 3,50 + 0,75(7) = 8,75
T2= 5,00 + 1,25(13) = 21,25
T3= 2,50 + 0,40(4) = 4,10
Total = 8,75 + 21,25 + 4,10 = 34,10
5. Entrada: S/. 15 (por persona)
Souvenir: S/. 5
Sea g: Gasto total por persona
x: Cantidad de souvenirs
Luego, g(x) = 15 + 5x
6.
2p = 2(3x + 7y) + 2(2x + 3y)
2p = 6x + 14y + 4x + 6y
2p = 10x + 20y
7. y(t) = 2,68 + 0,01t
y es una funcion afn
Para t = 8, tenemos:
y(8) = 2,68 + 0,01(8) = 2,76
4. 1 S/. (2x2+ x + 1)
2 S/. (x2+ x 3)
3 S/. (5x + 2)
4 S/. (4x + 1)
5 S/. y
Luego se cumple:
(2x
2
+ x + 1) + (x
2
+ x 3) + (5x + 2)+ (4x + 1) + y = 3x2+ 14x + 1
3x2+ 11x 1 + y = 3x2+ 14x + 1
y = 3x + 2
2x + 3y
3x + 7y
Clave: b
Clave: b
9. 1: 4x + 5
2: 3(4x + 5) = 12x + 15
3: 12x + 15 + 50 = 12x + 65
Total: 4x + 5 + 12x + 15 + 12x + 65
Total = 28x + 85
8. V = (4x2y) (7x3y2)(5x2 + y3)
V = 28x5y3(5x2 + y3)
V = 140x7y3+ 28x5y6
7/25/2019 Solucionario II
43/157
Matemtica II
Clave: a
Clave:c
Clave: b
Clave: e
Clave: a
Clave: c
Clave: c
Clave: e
Taller de prctica
1. A = {5; 6; 7; 8} n(A) = 4
B = {4; 5; 6; 7; 8} n(B) = 5
n(A B) = 4 x 5 = 20
3. E = 23 + 1923 8 1
= 4214
= 3
4. a + b = 8
a b = 2
2a = 10
a = 5 b = 3
Luego:
H = 9 3 + 2(5)
H = 16
5. 3a 1 = 5 b + 1 = 7
3a = 6 b = 6
a = 2
T1= 4x5y7 ; T2= 33x5y7
Piden: 4 + 33 = 29
6. P(x; y) = 6xm+1ym2+ 4xm+3ym1
2m 1 2m + 2
G.A.(P) = 2m + 2
18 = 2m + 2
m = 8
2. AR
B
.11 .5
.12 .6
.13 .7
.14 .8
.15 .9
.10
Ran(R) = {5; 6}
Pg. 88
10.y(t) = 1 500 + 55 t2
4
t: 1 dia = 24 h
y(24) = 1 500 + 5(24) 242
4
y(24) = 1 500 + 120 144
y(24) = 1 476
11.y = 72 + 6x x2
Para x = 10, tenemos:
y = 72 + 6(10) 102
y = 32
Clave: d
Clave: b
7. A(x) = 3x2+ 5x 7
B(x) = 2x2 4x + 1
A(x) + B(x) = x2+ x 6
8. Q(Q(x)) = Q(2x 5)
= 2(2x 5) 5
= 4x 10 5
= 4x 15
7/25/2019 Solucionario II
44/157
44 Matemtica II
Clave: e
9. y = 4 3xx + 5
yx + 5y = 4 3x
yx + 3x = 4 5y
x = 4 5yy + 3
x y + 3 0
y 3
Ran (f ) = {3}
1. f(x) = (3 (x + 1)2 1)(3 (x + 1)2+ 3x + 1 + 1)
3x(3 x + 1 + 1)
f(x) =(3x + 1
2 1)(3x + 1
2+ 3x + 1 + 1)
3x(3 x + 1 + 1)
f(x) = (3
x + 1 + 1)(3
x + 1 1) + (3
x + 12
+3
x + 1 + 13x(3 x + 1 + 1)
f(x) =(3x + 1 1)(3x + 12 + 3x + 1+ 1)
3x
f(x) =(3x + 1)3 13
3x
f(x) =x + 1 1
3x=
x3x
=13
Es decir, f es una funcin constante, pues ndepende de la variable "x".
Luego, f(1010) =13
Clave: b
Clave: d
Clave: a
Clave: a10. m + 2n = 13
2m n = 6 ...(2)
m + 2n = 13
4m 2n = 12
5m = 25
m = 5 n = 4
f = {(3; 13), (5; 6); (6; 11), (2; 3); (4; 5)}
Dom(f ) = {3; 5; 6; 2; 4}
Ran(f ) = {13; 6; 11; 3; 5}
Dom(f) Ran(f) = {3; 5; 6}
11.p + q r = 8 ...(I)
q + r p = 2 ...(II)
p q + 2 = 6 ...(III)
De (I) y (II): 2q = 6 q = 3
En (III) : p 3 + 2 = 6 p = 7
En (I): 7 + 3 r = 8 r = 2
Piden:
pq 1+ 7r + qr + 1= 72 7(2) + 33
= 49 14 + 27 = 62
12.b2+ 2b 35 = 0
b 7 b = 7
b 5 b = 5
3c 7 = 0 c =73
2d2 72 = 0 d2= 36 d = 6
Piden:
F = 5(4) 3(5)
9(73
) 6=
515
=13
Pg. 89
7/25/2019 Solucionario II
45/157
Matemtica II
Clave: c
2.
Si la grfica de la funcin cuadrtica no inter-secta al eje de las abscisas, entonces < 0, esdecir, b2 4ac < 0.
Anlisis y procedimiento:
Como la grfica de la funcin
f(x) = 9x2 6nx + (n + 12) no interseca al eje de
las abscisas, entonces sus races no son reales,luego:
< 0
(6n)2 4(9)(n + 12) < 0
36n2 36(n + 12) < 0
36(n2 n 12) < 0
n2 n 12 < 0
n 4
n +3
(n 4)(n +3) < 0
Por el mtodo de los puntos crticos se tiene
n 3; 4
Entonces los valores enteros de "n" son 2; 1;0, 1; 2; 3.
Piden: (2) + (1) + (0) + (1) + (2) + (3) = 3
+ +
x
f(x) = ax2+ bx + c; a > 0
y
3 4
Clave: c
Clave: b
3. Tenemos las funciones inyectivas
f(x) = ax2+ bx + c; x [2; +
g(x) = ax2+ bx + d; x ; 2]
con a 0 d y c no necesariamente diferente
Consideremos a > 0
d < c, entonces las grcas de f y g son:
4. As: P(x)x 3 3
R = P(3 3)
Luego, aplicamos la regla de Ruffini.
1 3 3 3 9 3 0 5 7
3 3 3 3 9 3 0 0 15
1 3 0 0 5 22
Como el residuo es R = 22 3
entonces P(3 3) = 22 3
x
vrtice
Resto
fg
2
y
En ambas grficas, el vrtice tiene abscisa h =
Como h =x1+ x2
22 =
b2a
4a = b
4a + b = 0
7/25/2019 Solucionario II
46/157
46 Matemtica II
Clave: e
Clave: cClave: d
Clave: a
Clave: c
Clave: c
Clave: e
Clave: e Clave: a
Clave: e
Clave: c
Clave: c
1. A = b h
A = (x + 1)2(x + 1)
A = 2 (x + 1)
2
A = 2x2+ 4x + 2
2. V = (x2+ 1)3
V = (x2)3+ 3(x2)(1) + 3(x2)(1)2+ 13
V = x6+ 3x4+ 3x2+ 1
3. A = b h
A = (x + 1)(x 1) = x
2
1
4. A = (x + 3)2+ (x 2)2+ 2x2+ 8
A = x2+ 6x + 9 + x2 4x + 4 + 2x2+ 8
A = 4x2+ 2x + 21
5. P = (x + 2)2 (2x + 1)2+ x
P = x2+ 4x + 4 (4x2+ 4x + 1) + x
P = x2+ 4x + 4 4x2 4x 1 + x
P = 3x2+ x + 3
7. G = (x + 8)(x + 4) (x + 4)(x + 5) 3(4x 3) + 9x
G = x2+ 12x + 32 (x2+ 9x + 20) 12x + 9 + 9x
G = x2+ 12x + 32 x2 9x 20 12x + 9 + 9x
G = 21
6. J = (x 7)(x + 7) (x 7)2+ x2+ 49
J = x2 49 (x2 14x + 49) + x2+ 49
J = x2 49 x2+ 14x 49 + x2+ 49
J = x2+ 14x 49
8. N = (a + 2)(a 2)(a2+ 22) + 16
N = (a2 4)(a2+ 4) + 16
N = a4 16 + 16
N = a4
9. M = (5x + 4)(4x + 5) 20(x + 1)2
M = 20x2+ 16x + 25x + 20 20(x2+ 2x + 1)
M = 20x2+ 41x + 20 20x2 40x 20
M = x
10.E = ( 7 + 2)2 ( 7 2)2
Aplicamos la identidad de Legendre.
E = 4( 7)( 2)
E = 4 14
11.E = 32 1 + 3(22+ 1)(24+ 1)(28+ 1)
E = 32 1 + (22 1)(22+ 1)(24+ 1) (28+ 1)
E = 32 1 + (24 1)(24+ 1)(28+ 1)
E =32
1 + (28
1)(28
+ 1)
E = 32 1 + 216 1
E = 32 216= 2
12.F =(x + 9)2 (x + 13)(x + 5)
(x + 10)(x + 9) (x + 16)(x + 3)
F =x2+ 18x + 81 (x2+ 18x + 65)
x2+ 19x + 90 (x2+ 19x + 48)
F = x
2
+ 18x + 81 x
2
18x 65x2+ 19x + 90 x2 19x 48
16
42=
8
21
4U N I D A D
Pg. 97
8
21
7/25/2019 Solucionario II
47/157
Matemtica II
Clave: c
Clave: c
Clave: c Clave: c
Clave: a
Clave: a
Clave: b
Clave: b
Clave: d
13.P = (x + 3)(x + 5) (x 2)(x 6)
P = x2+ 8x + 15 (x2 8x + 12)
P = x2+ 8x + 15 x2+ 8x 12
P = 16x + 3
14.A = (x + 2)2+ (x + 4)2 2(x + 3)2
A = x2+ 4x + 4 + x2+ 8x + 16 2(x2+ 6x + 9)
A = 2x2+ 12x + 20 2x2 12x 18
A = 2
15.R = (x + a)(x a)(x2+ a2)(x4+ a4) + a8
R = (x2 a2)(x2+ a2)(x4+ a4) + a8
R = (x4 a4)(x4+ a4) + a8
R = x8 a8+ a8
R = x8
16.M = 5(2 + 2)3 14(1 + 2)3
M = 5[23+ 3(2)2( 2) + 3(2)( 2)2+ ( 2)3]
14[13+ 3(1)2( 2) + 3(1)( 2)2+ 23]
M = 5[8 + 12 2 + 6(2) + 2 2] 14[1 + 3 2 +
3(2) + 2 2]
M = 5(8 + 14 2 + 12) 14(1 + 5 2 + 6)
M = 5(20 + 14 2) 14(7 + 5 2)
M = 100 + 70 2 98 70 2
M = 2
17.A = (x2+ x + 6)(x2+ x 3) (x2+ x + 9)(x2+ x 6)
Hacemos cambio de variable, y = x2+ x
A = (y + 6)(y 3) (y + 9)(y 6)
A = y2+ 3y 18 (y2+ 3y 54)
A = y2+ 3y 18 y2 3y + 54
A = 54 18
A = 36
18.K = (x + 2)(x2 2x + 4) (x 3)(x2 + 3x + 9)
K = x3+ 23 (x3 33)
K = x3+ 8 x3+ 27
K = 35
19.M =(x + 2a)2+ (2x a)2
(x + a)(x a) + 2a2
M =x2+ 4ax + 4a2+ 4x2 4ax + a2
x2 a2+ 2a2
M =5x2+ 5a2
x2+ a2
M =5(x2+ a2)
x2+ a2
M = 5
20.A = [(xn+ 3)2 6xn]2 x2n(x2n+ 18)
A = [x2n+ 6xn+ 9 6xn]2 x4n 18x2n
A = [x2n+ 9]2 x4n 18x2n
A = x4n+ 18x2n+ 81 x4n 18x2n
A = 81
21.N = ( 13 2 13 + 2 x2)(x2+ 3) 9
N = ( 132 22 x2)(x2+ 3) 9
N = ( 13 4 x2)(x2+ 3) 9
N = ( 9 x2)(x2+ 3) 9
N = (3 x2)(3 + x2) 9
N = 9 x4 9
N = x4
7/25/2019 Solucionario II
48/157
48 Matemtica II
22.[(x + a)(x b) + ab][(x a)(x + b) + ab]+ (ax bx)2
= [x2+ (a b)x ab + ab][x2+ (b a)x ab + ab]
+ (a b)2 x2
= [x2+ (a b)x][x2 (a b)x] + (a b)2 x2
= (x2)2 (a b)2 x2+ (a b)2 x2
= x4
26.Por dato: x + y = 3
Elevamos al cubo
(x + y)3= (3)3
x3+ 3xy(x + y) + y3= 27
x3+ y3= 27 3(3)(3)
x3+ y3= 0
27.Por dato: a + b = 5; ab = 2
Elevamos al cuadrado
(a + b)2= (5)2
a2+ 2ab + b2 = 25
a2+ b2+ 2(2) = 25
a2+ b2 = 21
28.x 35 = 2 x 2 = 35
Elevamos al cubo
(x 2)3= 353
x3 3(2)x2+ 3x(2)2 23= 5
x3 6x2+ 12x 8 = 5
P = 5
29.x2+ 3x = 2 x(x + 3) = 2
A = x(x + 3)(x + 1)(x + 2) 2 2
A = 2 (x2+ 3x + 2) 2 2
A = 2( 2 + 2) 2 2
A = 2
30.x y = 45 , xy = 5
Por Legendre:
(x + y)2 (x y)2= 4xy
(x + y)2 (45)2= 4 5
(x + y)2= 5 5
x + y = 5 5
23.P = 3 (37 + 1)(349 + 1 37)
P = 3 (37)3+ 13
P = 37 + 1
P = 38
P = 2
24.K = ( 3 + 2)2 + ( 3 2)2
K = 2( 32+ 22)
K = 2(3 + 2)
K = 10
25.R = ( 7 + 3)2+ ( 7 3)2
( 5 + 3)( 5 3)
R =2( 72 + 32)
52 32
R = 2(7+3)5 3
R = 10
Clave: d
Clave: a
Clave: a
Clave:
c
Clave: e
Clave: b
Clave: c
Clave: c
7/25/2019 Solucionario II
49/157
Matemtica II
31.M = (x + 3)(x + 4)(x + 2)(x + 5) (x 2+ 7x + 11)2
M = (x2+ 7x + 12)(x2+ 7x + 10) (x2+ 7x + 11)2
Sea x2+ 7x = a, luego:
M = (a + 12)(a + 10) (a + 11)2
M = a2+ 22a + 120 a2 22a 121
M = 1
34. n +1
n
2
= 3
n +1
n= 3
Ahora elevamos al cubo:
n +1
n
3
= ( 3)3
n3+1
n3+ 3(n)
1
n n +
1
n= 3 3
n3+1
n3+ 3 3 = 3 3
n3+1
n3= 0
35.x2+ y2+ z2 = 29 ; x + y + z = 9
(x + y + z) = 9, elevamos al cuadrado
(x + y + z)2= (9)2
x2+ y2+ z2+ 2(xy + xz + yz) = 81
29 + 2(xy + xz + yz) = 81
2(xy + xz + yz) = 52
xy + xz + yz = 26
36.a + b = 5; a2+ b2= 17
De Legendre:
(a + b)2+ (a b)2= 2(a2+ b2)
52+ (a b)2= 2(17)
(a b)2= 34 25
(a b)2= 9
a b = 3
37.x + y + z = 0
x3+ y3+ z3= 3xyz
x2
yz+
y2
xz+
z2
xy
=x3+ y3+ z3
xyz=
3xyz
xyz= 3
32.(x + 2y)2+ (x 2y)2= 8xy
Por Legendre:
2(x2+ 4y2) = 8xy
x2 4xy + 4y2= 0
(x 2y)2= 0 x = 2y
Piden: A =2(2y)(y) y2
(2y)2
A =3y2
4y2
A =3
4
33.Del dato ab = 1, en M
M =ab4+ a
b3+ a+
ba4+ b
a3+ b
M = (ab)b3+ a
b3+ a+ (ba)a
3+ b
a3+ b
M =b3+ a
b3+ a+
a3+ b
a3+ b
M = 2
Clave: b
Clave: b
Clave: c
Clave: a
Clave: c
Clave: a
Clave: aClave: d
x + y = 5 45
Piden:
x2 y2= (x + y)(x y)
= ( 5 45)(4 5)
= 5
7/25/2019 Solucionario II
50/157
50 Matemtica II
Clave: d
Clave: d
Clave: c
Clave: a
Clave: b
38.a + b = 6 ; a2+ b2= 30
De a + b = 6, elevamos al cuadrado
(a + b)2= (6)2
a2+ 2ab + b2= 36
a2+ b2+ 2ab = 36
30 + 2ab = 36
ab = 3
Piden:
a2
b+
b2
a=
a3 + b3
ab=
(a + b)(a2+ ab + b2)
ab
=6[(a2+ b2) + ab]
ab
=6[30 + 3]
3
= 66
39.R = (x2+ 5x + 5)2 (x + 1)(x + 2)(x + 3)(x + 4)
R = (x2+ 5x + 5)2 (x2+ 5x + 4)(x2+ 5x + 6)
Sea x2+ 5x = a, luego:
R = (a + 5)2 (a + 4)(a + 6)R = a2+ 10a + 25 a2 10a 24
R = 1
42.Por Legendre:
(a + b)2+ (a b)2= 2(a2+ b2)
(a + b)2
+ (a b)2
= 2(8)a + b = 16 (a b)2
Para que (a + b) sea mximo, entonces
a b = 0, luego
(a + b)max.= 4
41.x
2y+
2y
x= 2
x2 + 4y2
2xy= 2
x2+ 4y2= 4xy
x2 4y2= 4xy
(x 2y)2= 0
x 2y = 0
x = 2y x
y= 2
Piden: x
y
8
= (2)8= 256
40. a b = 2; b c = 2
Elevamos al cuadrado a b = 2
(a b)2= (2)2
a2 2ab + b2= 4 ... (I)
Elevamos al cuadrado b c = 2
(b c)2= (2)2
b2 2bc + c2= 4 ... (II)
Luego: a b = 2
b c = 2
a c = 4
Ahora elevamos al cuadrado a c = 4
(a c)2
= 42
a2 2ac + c2= 16 ... (III)
Sumamos (I), (II), (III)
a2 2ab + b2= 4
b2 2bc + c2= 4
a2 2ac + c2= 16
2a2+ 2b2+ 2c2 2ab 2bc 2ac = 24
2(a2+ b2+ c2 ab bc ac) = 24
a2+ b2+ c2 ab bc ac = 12
+
7/25/2019 Solucionario II
51/157
Matemtica II
Clave: a
Clave: e
Clave: c
Clave:e
Clave: e
44.xn
yn+
yn
xn= 62
x2n+ y2n= 62xnyn
x2n
+ y2n
+ 2xn
yn
= 64xn
yn
(xn+ yn)2= 64xnyn
xn+ yn
xn yn= 8
Reemplazamos en A:
A = 38 = 2
45.T = (nn nn)2+ (nn+ nn)2 2(n2n n2n)
T = 2(n2n+ n2n) 2(n2n n2n)
T = 4n2n
46. Por dato:
x2+ 3x + 1 = 0
Piden:
47. Nos piden: x2+ y2
ax + by = 8
ay bx = 6
a2+ b2 = 5
Elevamos al cuadrado ax + by = 8
(ax + by)2= (8)2
a2x2 + 2abxy + b2y2= 64(I)
(ay bx)2= 62
a2y2 2abxy + b2x2= 36(II)
Sumamos (I) y (II):
a2x2+ 2abxy + b2y2= 64
a2y2 2abxy + b2x2= 36
x2(a2+ b2) + y2(a2+ b2) = 100
(x2+ y2)(a2+ b2) = 100
(x2 + y2) 5 = 100
x2+ y2= 20
43.E = x2+1
x2 4(x +
1
x) + 6
E =x4+ 1
x2
4x2+ 4
x+ 6
E =x4+ 1 4x3 4x + 6x2
x2
E =x4 4x3+ 6x2 4x + 1
x2
D(x) = x4 4x3+ 6x2 4x + 1
Para x = 1 x 1 = 0
R(x) = D(1)
R(x) = 1 4 + 6 4 + 1
R(x) = 0
Entonces buscamos los factores primos con
Ruffini
1 4 6 4 1
x = 1 1 3 3 1
1 3 3 1 0
x = 1 1 2 1
1 2 1 0
Q(x) = x2 2x + 1 = (x 1)2
D(x) = (x 1)(x 1)(x 1)2= (x 1)4
E =(x 1)4
x2=
(x 1)2
x
x
x + 1+ =
= = = 0
1
x + 2
x2+ 2x + x + 1
x2+ 3x + 2
x2+ 3x + 1
x2+ 3x + 2
0
0 + 1
7/25/2019 Solucionario II
52/157
52 Matemtica II
Clave: b
Clave: b
Clave: eClave: e
Clave: e
Clave: c
Clave: d
48. K =(ax + by)2+ (ay bx)2
x2+ y2
K=a2x2+ 2abxy + b2y2+ a2y2 2abxy + b2x2
x2+ y2
K= x
2
(a
2
+ b
2
) + y
2
(a
2
+ b
2
)x2+ y2
K=(x2+ y2) (a2+ b2)
x2+ y2
K = a2+ b2
1. 16x4 8x2 12x3+ 8x 4x2 2
16x
4
+ 8x
2
4x
2
3x 12x3+ 8x
12x3 6x
2x
Q(x) = 4x2 3x
R(x) = 2x
2. 1 1 4 6 7 2
2 2 1
2 4 2
1 1 2 1
1 2 1 11 1
R(x) = 11x + 1
3. Completamos:
3x5+ 0x4 5x3+ 0x2 3x + 7
x2 x 1
1 3 0 5 0 3 7
1 3 3
1 3 3 3
1 1 1
4 4 4
3 3 1 4 2 11
5. Completamos y ordenamos:
1 3 2 1 0 0 2
0 0 3 3 1 2 0 2 2
1 4 0 4 4
1 0 1
3 2 4 1 2 3
Piden: Q(x) = 3x3+ 2x2+ 4x 1
6.
3 12 2 1 5 91 4 8
2 6 2 4
9 3 6
4 2 3 2 3
Piden: (2)(3) = 6
4. 8x
3
15x
4
+ 6x
5
23x
2
+ 42x3 1 5x2
Ordenamos:
6x5 15x4+ 8x3 23x2+ 0x + 4
2x3 5x2+ 0x 1
Por Horner:
2 6 15 8 23 0 4
5 15 0 3
0 0 0 0 01 8 20 0 4
3 0 4 0 0 8
Q(x) = 3x2+ 4
R(x) = 8
Pg. 104
Q(x) = 3x3+ 3x2+ x + 4
El trmino lineal es x
7/25/2019 Solucionario II
53/157
Matemtica II
Clave: e
Clave: c
Clave: b
Clave: b
Clave: bClave: c
7. Ordenamos:
x3 x2+ 33x + 20
x + 6
Por Ruffini:
x + 6 = 0 1 1 33 20
x = 6 6 30 18
1 5 3 2
Q(x) = x2+ 5x + 3
R(x) = 2
8. Ordenamos:
x3 x2+ 10x + 28
x + 6
Por Ruffini:
x + 6 = 0 1 1 10 28
x = 6 6 30 120
1 5 20 148
Q(x) = x2+ 5x 20
9.x3 2x + 1
x + 3
Por el teorema del resto:
x + 3 = 0
x = 3
R(x) = D(3)R(3) = (3)3 2(3) + 1
R(3) = 27 + 6 + 1
R(3) = 20
10.Por el teorema del resto:
x + 4 = 0
x = 4
R(x) = D(4)
R(x) = (4)4
+ 4(4)3
2(4)2
5(4) + 12R(x) = 44 44 32 + 20 + 12
R(x) = 0
11.Por el teorema del resto:
x + 1 = 0
x = 1
R(x) = D(1)
R(x) = 2(1)12+ (1)8+ (1)7+ 3(1)5 (1)
R(x) = 2 + 1 1 3 + 1 1
R(x) = 1
12.Por el teorema del resto:
5x 1 = 0
x = 15
R(x) = D1
5
R(x) = 151
5
4
81
5
3
91
5
2
+ 71
5+ 1
R(x) =15
625
8
125
9
25+
7
5+ 1
R(x) =15 40 225 + 875 + 625
625
R(x) =1 250
625= 2
R(x) = 2
7/25/2019 Solucionario II
54/157
54 Matemtica II
13. Por el teorema del resto:
x + 1 = 0
x = 1
R(x) = D(1)
R(x) = (1)5+ (1)4+ (1)3+ (1)2+ (1) + 1
R(x) = 1 + 1 1 + 1 1 + 1
R(x) = 0
16.Por el teorema del resto:
x a = 0
x = a
R(x) = P(a)
R(x) = a29+ 8a28+ 16a27
0 = a29+ 8a28+ 16a27
0 = a27(a2+ 8a + 16)
Como a 0
a2+ 8a + 16 = 0
(a + 4)2= 0
a = 4
17.Completamos:
3 3 0 7 A B
3 3 2
2 3 3 2
12 12 8
1 1 4 0 0
A 14 = 0 B + 8 = 0
A = 14 B = 8
Piden:A
B=
14
8=
7
4
18.Por el teorema del resto:
x 2 = 0
x = 2
R(x) = D(2)
R(x) = 25+ 2(2)4 3(2)3 2(2) a
14. 1 1 5 10 10 5 1
3 3 3 1
3 2 6 6 2
1 1 3 3 1
1 2 1 0 0 0
Q(x) = x2+ 2x + 1
15. 6x5+ 4x4 26x3+ 33x2 24x + 62x3+ 0x2 3x + 1
2 6 4 26 33 24 6 0 0 9 3
3 4 0 6 2
1 17 0 51
2
17
2
3 2 17
2 36 26
51
2 6+
17
2
Clave: c
Clave: e
Clave: a
Clave: b
Clave: c
Q(x) = 3x2+ 2x 17
2Piden:
3 + 2 17
2=
7
2
Clave: c
0 = 32 + 32 24 4 a
a = 64 28
a = 36
7/25/2019 Solucionario II
55/157
Matemtica II
25.
1 3 1 2 a a
1 3 3
1 4 4 4
9 9 93 4 9 0 b
a 13 = 0 ; a + 9 = b
a = 13 ; 13 + 9 = b
22 = b
23. Por el teorema del resto:
x + 2b = 0
x = 2b R(x) = (2b + 3b)7 [(2b)7 11b7]
R(x) = b7 (128b7 11b7)
R(x) = b7+ 139b7
R(x) = 140b7
24.
19.Por el teorema del resto:
x 1 = 0
x = 1
R(x) = D(1)
R(x) = [2(1) 1]16+ k(1)5+ 5(1)2 8
R(x) = [1 + k + 5 8]
0 = k 2
k = 2
R(x) = 62
3
3
192
3
2
+ 192
3 16
R(x) = 68
27 19
4
9+
38
3 16
R(x) =16
9
76
9+
38
3 16
R(x) =16 76 + 114 144
9
R(x) = 90
9
R(x) = 1020.1 2 1 0 a b
1 2 4
1 1 2
2 5 5 10
2 1 5 0 0
a + 7 = 0 10 b = 0
a = 7 10 = b
a b = 7 10
= 70
21.
5 5 11 10 a b
1 1 22 10 2 4
10 2 4
1 2 2 0 0
a 4 + 2 = 0 b + 4 = 0
a = 2 b = 4
Rpta.: 2 ; 4
Clave: a
Clave: a
Clave: e
Clave: a
Clave: c
Clave: c
Clave: d
x(x 1) + (x 2) (x + 1) + 4
(x + 2) (x 3)
Por el teorema del resto:x2 x 6 = 0
x2 x = 6
Reemplazamos en el dividendo:
R(x) = 6 + 6 + 2
= 14
(x2 x) + (x2 x) + 2
x2 x 6
22.Por el teorema del resto
3x 2 = 0
x =2
3
R(x) = D2
3
7/25/2019 Solucionario II
56/157
56 Matemtica II
Clave: e
Clave: b
Clave: a
Clave: b
26. R(x) = 5x2 3x + 7
2 8 0 4 m n p
1 4 0 12
0 4 2 0 6
3 2 1 0 3
4 2 1 5 3 7
m 11 = 5 n + 6 = 3 p + 3 = 7
m = 16 n = 9 p = 4
Piden:
m + n + p = 16 + (9) + 4 = 11
27.
3 6 0 1 a b
3 6 4
6 6 4
2 9 9 6
2 2 3 0 0
a + 13 = 0 ; b + 6 = 0
a = 13 b = 6
Piden: b a = 6 (13)
= 7
1.n + 2
4=
n + 4
5
5n + 10 = 4n + 16
n = 6
Clave: d
5. n = 7 ; k = 4
tk= xn k , yk 1
t4= x3 y3
Clave: a
4. 2p
2= 3q
1= 12
p = 12 ; q = 4
Piden: p + q = 16
Clave: c
6.155
5= 31 ;
93
3= 31 n = 31
tk= (1)k 1 xn k , yk 1 , k = 3
t3= x28 y2
Clave: e
7. n = 25 tc =
n + 1
2 = 13
t13= (x4)12 (y4)12
t13= x48 y48
2.xn+ ym
x3+ y4
n
3=
m
4= 21
n = 3(21) = 63
m = 4(21) = 84
Pg. 110
m + n = 84 + 63
m + n = 147
Clave: c
3.3a + 6
3=
a + 12
2
6a + 12 = 3a + 36
3a = 24 a = 8
N trminos = 3(8) + 63
= 10
7/25/2019 Solucionario II
57/157
Matemtica II
Clave: a
8.(3x)4 (1)4
3x 1n = 4 ; k = 3
t3= (3x)1(1)2
t3= 3x
Clave: b
9.(x + 5)5 25
(x + 5)1 21 n = 5 ; k = 4
t4= (x + 5)1 (2)3
t4= (x + 5)8
t4= 8x + 40
Clave: a
12. (x2
)41
141
x2 1
n = 41 ; k = 3
t3= (x2)41 3(1)3 1
t3= x76
Clave: b
14.a75 b30
a15 b6=
(a15)5 (b6)5
a15 b6
n = 5 ; k = 5
t5= (a15)5 5 (b6)5 1
t5= b24
G.A. = 24
Clave: b
16.(x2)2 (3xy2)2
x2 (3xy2)
x3+ (2y2)3
x + 2y2
= x2+ 3xy2 (x2 x(2y2) + (2y2)2)
= x2+ 3xy2 x2+ 2xy2 4y4
= 5xy2 4y4
Clave: e
13.(2x)5 y5
2x yn = 5
tc : k =n + 1
2=
5 + 1
2= 3
t3= (2x)5 3y3 1
t3= 4x2y2
Clave: d
10.y8 x8
y + x
n = 8 ; k = 6
tk= (1)k 1yn kxk 1
t6= y8 6x6 1
t6= y2x5
Clave: e
11.(x3)7+ (y3)7
x3+ y3
n = 7 ; k = 5
t5= (1)5 1 (x3)7 5 (y3)5 1
t5= x6y12
Clave: a
15. x3+ 1x + 1
+ x 1
x2 x + 1 + x 1 = x2
7/25/2019 Solucionario II
58/157
58 Matemtica II
17.x 3x 81
3x 3
x
1 + 34
x 3
=x 34
x 3
n = 4 y k = 3
t3= x4 3
(3)3 1
t3= 9x
13
13
13
13
43
13
Clave: c
Clave: e
Clave: c
Clave: a
Clave: e Clave: e
18.x64 y48
x4 y3
=(x4)16 (y3)16
x4 y3
n = 16 ; k = 6
t6= (x4)16 6 (y3)6 1
t6= x40y15
G.A. = 40 + 15G.A. = 55
19.a38 b57 c19
a2 b3 c
(a2)19 (b3c)19
a2 b3c
n = 19 ; k = 10
t10= (a2)19 10(b3c)10 1
t10= a18b27c9
G.A. = 18 + 27 + 9
G.A. = 54
20.(x + 3)36 x36
3
=(x + 3)36 x36
(x + 3) x
n = 36 ; k = 29
t29= (x + 3)36 29 (x)29 1
Para: x = 1:
t29= (1 + 3)7 (1)28
t29= 128
21. (x + 4)3 64x
=(x + 4)3 43
(x + 4) 4
n = 3 ; k = 2
t2= (x + 4)3 2 (4)2 1
t2= 4(x + 4)
22.P =x102+ x96+ x90+ + 1
x90+ x72+ x54+ + 1
P =
(x6)18 118
x6 1
=
x108 1
x6 1
P = x18
1x6 1
= (x6
)3
13
x6 1
P = (x6)2+ x6+ 1
P = x12+ x6+ 1
(x18)6 16
x18 1
x108 1
x18 1
x
4
34
x 3
13
13
7/25/2019 Solucionario II
59/157
Matemtica II
25.x75 yb
xc y2
75
c=
b
2= n
75
c= n
b
2= n
tk= (xc)
k
(y2)k 1
xay24= x(75 kc)y(2k 2)
a = 75 kc 24 = 2k 2
13 = k
Por termino central:
13 =n + 1
2
25 = n
Clave: e
Clave: b
75c
Clave:a
Clave: a
Clave: e
24.xa yb
x3 y7, se cumple:
a
3=
b
7= 9
a = 27 ; b = 63
(x3)9 (y7)9
x3 y7
t6= (x3)
9 6 (y7)6 1
t6= x9y35
26.x
4n 2
x
4n 4
+ x
4n 6
+ x
2
1
=x4n + 1
x2+ 1
27.xa + yb
x + y2 ; Por dato: tk= x
4y10
a
1=
b
2a = r ; b = 2r
xr + (y2)r
x + y2
tk= (1)k 1xr k (y2)k 1= x4y10
Se cumple:
r k = 4 ; 2k 2 = 10
k = 6
r = 6 + 4
r = 10
Piden:
a b = r 2r
= 2r2
= 2(10)2
= 200
23.xm yn
x5 y7 ;
m
5=
n
7m = 5r ; n = 7r
x5r y7r
x5 y7=
(x5)r (y7)r
x5 y7
t16= (x5)r 16 (y7)16 1
x95y105= x5r 80y105
Luego, se cumple:
5r 80 = 95
5r = 175
r = 35
Piden:
m + n = 5r + 7r
= 12r
= 12(35)
= 420
Luego:75
c= 25 c = 3
b
2= 25 b = 50
De: a = 75 kc
a = 75 (13)(3)
a = 36
Piden:
a + b + c = 36 + 3 + 50
a + b + c = 89
7/25/2019 Solucionario II
60/157
60 Matemtica II
28.xa yb
x3 y4
a
3=
b
4a= 3r ; b= 4r
t6= (x3
)r 6
(y4
)6 1
t6= x3r 18y20
t9= (x3)r 9 (y4)9 1
t9= x3r 27y32
t7= (x3)r 7 (y4)7 1
t7= x3r 21y24
Por dato:
t6 t9
t7= x12y28
x3r 18y20 x3r 27y32x3r 21y24
= x12y28
x3r 18 + 3r 27 3r + 21 y20 + 32 24= x12y28
x3r 24y28 = x12y28
3r 24 = 12
r = 12
Piden:
a+ b= 7r
= 7(12) = 84
Clave: b
Clave: d
Clave: e
Clave: c
Clave: e
Clave: c
Clave: a
Miscelnea Pg. 112 y 113
1. Por el teorema del resto:
x 1 = 0
x = 1
R(x) = D(1)
R(x) = (1)3 2(1) + 1
R(x) = 0
2. (x + 8)(x 8) = x2 64
3. (x 3)
3
= x
3
3(x
2
)(3) + 3x(3)
2
3
3
= x3 9x2+ 27x 27
4. Como contamos a partir del final:
x3n yn
x3 y=
yn x3n
y x3
t8= (y)n 8 (x3)8 1
t8= yn 8x21
G.A. = n 8 + 21
38 = n + 13
25 = n
5. a + b = 6 ; ab = 3
Elevamos al cubo: a + b = 6
(a + b)3= 63
a3+ b3+ 3ab(a + b) = 216
a3+ b3+ 3(3)(6) = 216 a3+ b3= 162
6. A = pr2
A = p(2x + 1)2
A = p(4x2+ 4x + 1)
7.xm yn
x5 y7 ;
m
5=
n
7 m = 5r ; n = 7r
t8= (x5)r 8 (y7)8 1
t8= x5r 40y49
7/25/2019 Solucionario II
61/157
Matemtica II
Clave: e
Clave: e
Clave: d
Clave: e
Clave: b
Clave: a
8. a + b = 3 ; ab = 4
Elevamos al cuadrado a + b = 3
(a + b)2= (3)2
a2+ 2ab + b2= 9
a2+ b2+ 2(4) = 9
a2+ b2= 1
10.A=b h
2
A= (x 3)(x2
+ 3x + 9)2
A=x3 33
2
A=x3 27
2
12.
Averde= (x + 2)2+ 4 2 + (x + 1)2
= x2+ 4x + 4 + 8 + x2+ 2x + 1
= 2x2+ 6x + 13
11.
2 a 4 e 11 q r
1 = b 2 c = 6 4
3 d = 2 1 3 2 2 6 n = 3 9 6
2 m = 1 3 p = 1 3 10
Luego:
A = (m + n + p + q + r) (a + b + c + d + e)
A = (1 + 3 1 + 8 + 4) (4 + 1 6 2 + 13)
A = 13 10 = 3
9. A = pr2
D = (x 2)(x + 2)
D = (x2 4)
r =x2 4
2
Nos piden el rea:
A = px2 4
2
2
A = px2 8x + 16
4
G.A. = 5r 40 + 49
59 = 5r + 9
10 = r
Piden:
m + n = 5r + 7r
= 12r
= 12(10)
= 120
= 13: = 8 = 4=
4
x + 1
x + 3
x + 6
x + 2
2x + 7
3x + 10
4x
2
x + 1
x + 2
x + 1
2x + 3
13.Aazul= (x + 3)(3x + 4) + (x + 6)(x + 1)
= 3x2+ 13x + 12 + x2+ 7x + 6
= 4x2+ 20x + 18
Clave: c
7/25/2019 Solucionario II
62/157
62 Matemtica II
Clave: e
Clave: b
Clave: a
Clave: c
14.
AT= 2[(x + 4)(x + 2) + (x + 4)(x + 1) + (x + 2)(x + 1)]
AT= 2(x2+ 6x + 8 + x2+ 5x + 4 + x2+ 3x + 2)
AT= 2(3x2+ 14x + 14)
AT= 6x2+ 28x + 28
x + 4
x + 1
x + 2
Clave: e
Clave: e
Clave: e
Clave: e
Pg. 115
1. P = (x + 2)(x 2) (x 1)2 6x 1
P = x2 4 (x2 2x + 1) 6x 1
P = x2 4 x2+ 2x 1 6x 1
P = 4x 6
15.V = (x + 4)(x + 2)(x + 1)
V = (x2+ 6x + 8)(x + 1)
V = x3 + 7x2+ 14x + 8
4. E = ( 5 + 2)2+ ( 5 2)2
E = 2[( 5)2+ ( 2)2]
E = 2(5 + 2)
E = 14
Clave: b
6. Por el teorema del resto:
x + 2 = 0
x = 2
R(x) = D(2)
R(2) = 3(2)3+ (2)2 5(2) + 20
R(2) = 24 + 4 + 10 + 20
R(2) = 10
Clave: e
5. a2+ b2= 3ab
(a + b)2 (a b)2
(a + b)2+ (a b)2
4ab
2(a2+ b2)=
4ab
2(3ab)=
2
3
7. Por el teorema del resto:
x 1 = 0 x = 1
R(x) = D(1)
R(x) = (1)5+ (1)4+ (1)3 (1)2+ 6
R(x) = 8
8. Por el teorema del resto:
x 1 = 0 ; R(x) = 0
x = 1
R(x) = D(1)
R(1) = (1)3+ m(1)2+ n(1) + 1
0 = 1 + m + n + 1
2 = m + n
2. M = (x 3)(x + 3) + (x 2) 2+ 2x2+ 2x + 13
M = x2 9 + x2 4x + 4 + 2x2+ 2x + 13
M = 4x2 2x + 8
3. A = 2x + 13 + (2x + 5)2 (2x 5)2
A = 2x + 13 + 4(2x)(5)
A = 2x + 13 + 40x
A = 42x + 13
7/25/2019 Solucionario II
63/157
Matemtica II
12.( 2 + 3 + 5)( 2 + 3 5)
= [( 2 + 3) + 5)][ 2 + 3) 5]
Por diferencia de cuadrados.
= ( 2 + 3)2
52
= 22+ 2 2 3 + 32 5
= 2 + 2 6 + 3 5
= 2 6
13.
1 6 b 12 e
2 a 8 c 4
1 4 5 d 0
e 4 = 0 ; 2d = 4 ; 12 + c = 2
e = 4 d = 2 c = 10
b 8 = 5 6 + a = 4
b = 13 a = 2
Nos piden:
a + b c + d e = 2 + 13 (10) + 2 4
a + b c + d e = 19
Clave: d
Clave: d
10.x5 32
x 2
=(x)5 (2)5
x 2 ; n = 5 ; k = 3
t3= x5 323 1
t3= 4x2
Clave: e
14.32x5 + 243y5
2x + 3y
=(2x)5 (3y)5
2x+ 3y
t4= (1)41(2x)5 4(3y)4 1
t4= (2x)(3y)3 t4= 54xy
3
Clave: e
11.x35 y49
x5 y7
=(x5)7 (y7)7
x5
y7 ; n = 7
El trmino central es t4, luego:
t4= (x5)7 4 (y7)4 1
t4= x15y21
Clave: e
Clave: e
9.x2 5x + 6 + 25
x2 5x + 1
x2 5x + 31
x2 5x + 1
y + 31
y + 1
y
y
Por el teorema del resto:
y + 1 = 0 y = 1
R(y) = D( 1)
= 1 + 31
= 30
R(x) = 30
7/25/2019 Solucionario II
64/157
64 Matemtica II
Clave: e
Clave:b
Clave: a
Clave: d
1. M =
[(a + b)2+ (a b)2][(a + b)2 (a b)2]
2a2+ 2b2
M =2(a2+ b2) (4ab)
2(a2+ b2)
M = 4ab
2. E =an
bn+
bn
an=
a2n+ b2n
an bn ... (I)
Por dato:
an+ bn= 3 anbn
Elevamos al cuadrado
(an+ bn)2= (3 an bn)2
a2n+ 2anbn+ b2n= 9anbn
a2n+ b2n= 7anbn
Reemplazamos en (I)
E =a2n+ b2n
a
n
b
n =7anbn
a
n
b
n = 7
3. x3y4+ x2y2= x2y2(xy2+ 1) = 3xy
xy xy2+ 1 = 3xy
Cancelamos xy en ambos miembros y elevam
al cuadrado.
xy2+ 1 = 9 xy2= 8
Luego: 3xy2= 38 = 2
4. xm+1
xm= 2
x2m+ 1 = 2xm
x2m 2xm+ 1 = 0
(xm 1)2= 0
Luego: xm= 1
Piden: x3m+ x-3m
= (xm)3 + (xm)3
= 13+ 13= 2
Pg. 116
7/25/2019 Solucionario II
65/157
Matemtica II
Clave: d
Clave: a
Clave: c
Clave: c
Clave: bClave: e
Clave: e
Clave: e
Clave: e
Clave: e
Clave: a
1. P(x) = (x + 3)(x2 2x + 1)
P(x) = (x + 3)(x 1)2
Luego un factor es (x 1)
8. F(a; b) = 5a9b3+ 15a6b7
F(a; b) = 5a6b3 (a3+ 3a4)
3 factores primos
9. T(a; b) = a3+ a2b + ab2+ b3
T(a; b) = (a3+ ab2) + (a2b + b3)
T(a; b) = a(a2+ b2) + b(a2+ b2)
T(a; b) = (a2+ b2)(a + b)
2. 2bx3 3abx2+ 2cx 3ac
bx2(2x 3a) + c(2x 3a)
(2x 3a)(bx2+ c)
2 factores primos
3. Es directa por producto notable.
4. P(x) = ax4 bx3+ x2
P(x) = x2 (ax2 bx + 1)
2 factores primos
5. P(x) = a(x 3) x + 3
P(x) = a(x 3) (x 3)
P(x) = (x 3)(a 1)
6. P(x) = x3+ 8
P(x) = x3+ 23
P(x) = (x + 2)(x2 2x + 4)
Calculamos la suma de los trminos indepen-dientes:
2 + 4 = 6
7.
P(x; y) = y
2
n
2
+ x
2
m
2
+ y
2
m
2
+ x
2
n
2
Agrupamos:
(y2n2+ y2m2) + (x2m2+ x2n2)
y2 (n2+ m2) + x2(m2+ n2)
(n2+ m2)(x2 + y2)
10. P(x) = x3+ x2 x 1
Agrupamos:
P(x) = (x3 x) + (x2 1)P(x) = x(x2 1) + (x2 1)
P(x) = (x2 1)(x + 1)
P(x) = (x + 1)(x 1)(x + 1)
P(x) = (x + 1)2 (x 1)
11. a(1 b2) + b(1 a2)
a ab2+ b a2b
(a a2b) + (ab2+ b)
a(1 ab) + b(ab + 1)
a(1 ab) + b(1 ab)
(1 ab)(a + b)
12. P(x; y) = (x + 1)2 (y 2)2
P(x; y) = [(x + 1) + (y 2)] [(x + 1) (y 2)]
P(x; y) = (x 1 + y)(x + 3 y)
13. R(x) = 8x3+ 27
= (2x)3+ 33
= (2x + 3)[(2x)2 (2x)(3) + 32]
5U N I D A D
Pg. 122
7/25/2019 Solucionario II
66/157
66 Matemtica II
Clave: c
Clave: c
Clave: a
Clave: d
Clave: d
Clave:
e
Clave: c Clave: c
14.T(x; y) = (xy + 1)2 (x + y)2
T(x; y) = (xy)2+ 2xy + 1 x2 2xy y2
T(x; y) = (xy)2+ 1 x2 y2
= (x2y2 x2) (y2 1)
= x2(y2 1) (y2 1)
= (y2 1) (x2 1)
= (y + 1)(y 1)(x + 1)(x 1)
Calculamos la suma de los factores primos:= y + 1 + y 1 + x + 1 + x 1
= 2y + 2x
= 2(x + y)
17. P(x) = x7+ c3x4 c4x3 c7
Agrupamos:
(x7+ c3x4) (c4x3+ c7)
x4 (x3+ c3) c4(x3+ c3)
(x3+ c3)(x4 c4)
(x3+ c3)(x2+ c2)(x2 c2)
(x3+ c3)(x2+ c2)(x + c)(x c)
(x + c)(x2 xc + c2)(x2+ c2)(x + c)(x c)
5 factores primos
18.
P(x; y) = a
2
b
2
+ x
2
y
2
+ 2(ax by)P(x; y) = a2 b2+ x2 y2+ 2ax 2by
P(x; y) = (a2+ 2ax + y2) (b2+ 2by + y2)
P(x; y) = (a + x)2 (b + y)2
P(x; y) = (a + x + b + y)(a + x b y)
19. 4x4+ 81y4+ 36x2y2
4x4+ 36x2y2+ 81y4
(2x2)2+ 36(xy)2+ (9y2)2
(2x2+ 9y2)2
20. E(x) = (x 3)(x 2)(x 1) + (x + 2)(x 1) (x 1
E(x) = (x 1)[(x 3)(x 2) + x + 2 1]
E(x) = (x 1)(x2 5x + 6 + x + 1)
E(x) = (x 1)(x2 4x + 7)
Factor primo lineal: x 1
15. P(x) = x3 125
P(x) = x3 53
P(x) = (x 5)(x2+ 5x + 25)
2 factores primos
16. P(a) = a3+ 2a2 a 2
Los divisores del T.I. 2: {1; 2}
1 2 1 2
a = 1 1 3 2
1 3 2 0
(a 1)(a2+ 3a + 2)
(a 2) (a 1)
(a 1)(a + 2)(a + 1)
El factor primo con mayor trmino indepen-diente es (a + 2).
= (2x + 3)(4x2 6x + 9)
F.P.: 2x + 3 la suma de coeficientes es 2 + 3 = 5
F.P.: 4x2 6x + 9 la suma de coeficientes es
4 6 + 9 = 7
7/25/2019 Solucionario II
67/157
Matemtica II
21. P(a; b; c) = a(b2+ c2) + b(c2+ a2)
P(a; b; c) = ab2+ ac2+ bc2+ ba2
P(a; b; c) = (ab2+ ba2) + (ac2+ bc2)
P(a; b; c) = ab(b + a) + c2(a + b)
P(a; b; c) = (a + b) (ab + c2)
Suma de
coeficientes
22. x2 5xy 14y2 41y + 2x 15
x 2y 5
x 7y 3
(x + 2y + 5)(x 7y 3)
24. P(x) = 4x2 4x 3
2x 3
2x +1
P(x) = (2x 3)(2x + 1)
Suma de coeficientes 2 3 = 1
2 + 1 = 3
25. x2+ 2x 120
x +12
x 10
(x + 12)(x 10)
Suma de factores primos: 2x + 2
26. 6x2 13xy + 2y2+ 5y 8x + 2
6x y 2
x 2y 1
(6x y 2)(x 2y 1)
27. 2x2+ 3y2+ 7xy y + 3x 2
Ordenamos:
2x2+ 7xy + 3y2 y + 3x 2
2x 1y 1
x 3y 2
(2x + y 1)(x + 3y + 2)
28. x2 y2+ 10y 25x2+ 0xy y2+ 0x + 10y 25
x y 5
x y 5
(x y + 5)(x + y 5)
29. x3 4x2 13x 8
Posibles ceros: {1, 2, 4, 8}
1 4 13 8
x = 1 1 5 8
1 5 8 0
(x + 1)(x2 5x 8)
= 1 + 1 = 2
Clave: b
Clave: e
Clave: a
Clave: b
Clave: a
Clave: a
Clave: b
Clave: c
Clave: b
23. x3 2x2+ 3x + 6
Los divisores del trmino independiente
6: (1, 2, 3, 6)
1 2 3 6
x = 1 1 3 6 1 3 6 0
(x + 1)(x2 3x + 6)
7/25/2019 Solucionario II
68/157
68 Matemtica II
30. x3 14x2+ 47x + 8
Posibles ceros: {1, 2, 4, 8}
1 14 47 8
x = 8 8 48 8
1 6 1 0
(x 8)(x2 6x 1)
31. P(a; b) = a(b2+ b + 1) + b(a2+ a + 1) + a2+ b2
P(a; b) = ab2+ ab + a + a2b + ab + b + a2+ b2
P(a; b) = (a2b + ab2) + (a + b) + (a2+ 2ab + b2)
P(a; b) = ab(a + b) + (a + b) + (a + b)2
P(a; b) = (a + b) [(a(1 + b) + (1 + b)]
P(a; b) = (a + b)(1 + b)(a + 1)
36.mn4 5m2n3 4m3n2+ 20m4n
(mn4 5m2n3) (4m2n2 20m4n)
mn3 (n 5m) 4m3n (n 5m)
(n 5m)(mn3 4m3n)
mn(n 5m)(n2 4m2)
mn(n 5m)(n + 2m)(n 2m)
37. P(x; y; z) = x2+ y2+ x(y + z) + y(x + z)
P(x; y; z) = x2+ y2+ xy + zx + yx + zy
P(x; y; z) = (x2+ xy) + (y2+ yx) + (zx + zy)
P(x; y; z) = x(x + y) + y(y + x) + z(x + y)
P(x; y; z) = (x + y)(x + y + z)
38. P(x) = x20 (x27+ x20+ 1) + x7(x20+ 1) + 1
P(x) = x47+ x40+ x20+ x27+ x7+ 1
P(x) = (x47+ x40) + (x27+ x20) + x7+ 1
P(x) = x40(x7+ 1) + x20(x7+ 1) + (x7+ 1)
P(x) = (x7+ 1) (x40 + x20+ 1)
32.A(x) = x4+ 6x2+ 9
x2 3
x2 3
(x2+ 3)2
33. P(x) = 4x4+ 26x2+ 36
2x2 4
2x2 9
(2x2+ 4)(2x2+ 9)
34. x2+ 6y2 5xy x 6
x2 5xy + 6y2 x + y 6
x 3y 3
x 2y +2
(x 3y 3)(x 2y + 2)
35.M(x) = x3+ x2 x 1
Posibles ceros = {1}
1 1 1 1
x = 1 1 2 1
1 2 1 0
(x 1)(x2+ 2x + 1)Clave: c
Clave: d
Clave: d
Clave: e
Clave: a
Clave: a
Clave: d
Clave: b
Clave: c
(x 1) (x + 1)2
Piden:
1 1 = 0 1 + 1 = 2
7/25/2019 Solucionario II
69/157
Matemtica II
40. x2 4xy + 3y2 8y + 4x + 4
x 3y 2
x y 2
(x 3y + 2)(x y + 2)
41. x3+ 2x2 4x 8
Posibles ceros: {1, 2, 4, 8}
1 2 4 8
x = 2 2 8 8
1 4 4 0
(x 2)(x2+ 4x + 4) = (x 2)(x + 2) 2
Rpta.: x + 2
Clave: e
Clave: c
Clave: d
39. x3+ 6x2+ 14x + 15
Posibles ceros: {1, 3, 5, 15}
1 6 14 15
x = 3 3 9 15
1 3 5 0
(x + 3)(x2+ 3x + 5)
Clave: c
Clave: a
1. Sea T la longitud inicial de la tela
Se vende:34
T
Queda: 1
4T
Luego 14
T = 30
T = 120
2. Guillermo en 1 dia har1
12de la obra
Guillermo y Jos en 1 dia harn18
de la obra
Jos en un da har:
18
1
12=
12 896
=4
96=
124
Luego:
En 1 da1
24obra
En x das 1 obra
(toda la obra)
x =1
24
1 1
x = 24
3. Seaab
la fraccin, luego:
a + 5b + 5
=45
5(a + 5) = 4(b + 5)
5a + 25 = 4b + 20
5a 4b = 5 (I)
a 2b 2
=58
8(a 2) = 5(b 2)
8a 16 = 5b 10
8a 5b = 6 (II)
Multiplicamos (I) por 5 y (II) por 4
(5a 4b = 5) (5)
(8a 5b = 6) (4)
25a + 20b = 25(+)
32a 20b = 24
7a = 49
a = 7
Pg. 129
7/25/2019 Solucionario II
70/157
70 Matemtica II
Clave: b
Clave: b
Clave: b
Clave: d
Clave: a
Clave: a
Clave: a
4. 4 (3x 2) = 12
4 3x + 2 = 12
6 3x = 12
x = 2
8.3 x + 7 = 93 x = 2
x = 8
9. 3x 4y = 41 (I)
11x + 6y = 47 (II)
Multiplicamos (I) por 6 y (II) por 4
(3x 4y = 41) (6)
(11x + 6y = 47) (4)
18x 24y = 246+
44x + 24y = 188
62x = 434 x = 7
En (I): 3(7) 4y = 41
21 4y = 41
4y = 20
y = 5
Luego: xy = 35
5. 3x 4(x + 3) = 8x + 6
3x 4x 12 = 8x + 6
x 12 = 8x + 6
9x = 18
x = 2
6. 8x 15x 30x 51x = 53 + 31x 172
88x = 119 + 31x
119x = 119
x = 1
7. (5 3x) (4x + 6) = (8x + 11) (3x 6)
5 3x + 4x 6 = 8x + 11 3x + 6
1 + x = 5x + 17 4x = 18
x = 92
En (I): 5(7) 4b = 5
35 4b = 5
4b = 40
b = 10
Piden: a + b = 17
10. 9x + 11y = 14 .(I)
6x 5y = 34 .(II)
Multiplicamos (I) por 5 y (II) por 11
(9x + 11y = 14) (5)
(6x 5y = 34) (11)
45x + 55y = 70+
66x 55y = 374
111x = 444
x = 4
En (I): 9(4) + 11y = 14
36 + 11y = 14
11y = 22
7/25/2019 Solucionario II
71/157
Matemtica II
y = 2
Luego: x + 2y = 0
Clave: e
Clave: a
Clave: b
Clave: a
Clave: e
11. 10x 3y = 36 (I)
2x + 5y = 4 .(II)
Multiplicamos (I) por 5 y (II) por 3
(10x 3y = 36) (5)
(2x + 5y = 4) (3)
50x 15y = 180+
6x + 15y = 12
56x = 168
x = 3
En (I): 10(3) 3y = 36
30 3y = 36
y = 2
Luego: x = 3 ; y = 2
Piden:
x y = 3 (2) = 5
13.x6
+ 2x + 1 = 5x12
34
MCM(6; 12; 4) = 12
Multiplicamos por 12 a cada una de los trmino
12 x6
+ 12 2x + 12 1 = 12 5x12
12 34
2x + 24x + 12 = 5x 9
21x = 21
x = 1
14.4x9
13
= 2x + 15
MCM (9; 3; 5) = 45
Multiplicamos por 45 a cada uno de los trmnos
45 4x9
45 13
= 45 2x + 15
5 4x 15 = 9 (2x + 1)
20x 15 = 18x + 9
x = 12
15.12x +
14 =
110x+
15
MCM (2; 4; 10; 5) = 20
Multiplicamos por 20 a cada uno de los trmnos
12.(m 3)
2 (2m + 1)
3+ m
5+ 3 = 2m 1
3
MCM (2; 3; 5) = 30
Multiplicamos por 30 a cada uno de los trminos
30 (m 3)2
30 (2m + 1)3
+ 30 m5
+ 30 2m
= 30 2m 30 13
15(m 3) 10(2m + 1) + 6m + 90 = 60m 10
15m 45 20m 10 + 6m + 90 = 60m 10
m + 35 = 60m 10
59m = 45
m = 4559
7/25/2019 Solucionario II
72/157
72 Matemtica II
22.a2x + b2= b2x + a2+ a b
x(a2 b2) = a2 b2+ a b
x =(a + b)(a b) + a b
(a + b)(a b)
x =(a b)(a + b + 1)
(a + b)(a b)
Clave: c
Clave: e
Clave: a
Clave: b
Clave: e
Clave: e
Clave: c
16. 3x x3
= 48
8x3
= 48
x = 18
18.
Falta Sobra
Nmero de hijos =10 + 48 6
= 7
Nmero de caramelos = 7 6 + 4 = 46
19.
Falta Sobra
Nmero de carpetas =4 + 12 1
= 5
Nmero de alumnos = 5 1 + 4 = 9
20. 5x 6y = 60
7x + 3y = 27 (2)
5x 6y = 60(+)
14x + 6y = 54
19x = 114
x = 6
21. MCM = 15
15x + 9x 27 = 60 25x + 60
24x 27 = 120 25x
49x = 147
x = 3
17. Sean D y d los nmeros:
D nmero mayor
d nmero menor
D d D = 3d + 9 (I)
9 3
d + 300 D72 3 1
d + 300 = (3 1) D + 72
d + 300 = 2D + 72
Luego de (I):
d + 300 = 2(3d + 9) + 72
d + 300 = 6d + 18 + 72
5d = 210
d = 42De (I): D = 3(42) + 9
D = 135
Luego: D = 135, d = 42
20 12x
+ 20 14
= 20 110x
+ 20 15
10x
+ 5 = 2x
+ 4
1 = 8x
x = 8
Luego: 3 x = 38 = 2
7/25/2019 Solucionario II
73/157
Matemtica II
23. Sobra Sobra
Nmero de hijos =37 11
8 6= 13
Nmero de caramelos = 13 8 + 11 = 115
Suma de cifras: 1 + 1 + 5 = 7
27.(I) 3x 2y = 12 (4)
(II) 15x + 8y = 20
12x 8y = 48(+)
15x + 8y = 20
27x = 28 x =
28
27
Reemplazamos en (I):
32827
2y = 12
289
12 = 2y
809
= 2y y = 409
Piden:
x + y =2827
409
=28 120
27=
9227
24.Si Rosa tiene S/. 80 y Mara tiene S/. 60, enton-ces lo que han ganado es:
340 80 60 = 200Luego cada una gan = 100
Nmero de das =10020
= 5
25. Pedro: x
Juana: 4x + 20
Luego: x + 4x + 20 = 920
x = 180 Pedro tiene S/. 180
26.Prez: P
Quispe: Q
P Q = 38 (I)
P 7 = 2Q
P = 2Q + 7 (II)
Reemplazamos (II) en (I):2Q + 7 Q = 38
Q = 31
Luego el seor Quispe tiene 31 aos
Clave: a
Clave: c
Clave: e
Clave: b
Clave: b
Clave: d
Clave: e
1.
Sea m el nmero de manzanas:3m 1 < m + 3
2m < 4
m < 2
Entonces:
m = 1
Pg. 133
x =a + b + 1
a + b
Clave: e
2. x: N de canicas
x + 8 > 18 x > 10 (I)x 5 < 7 x < 12 (II)
De (I) y (II):
10 < x < 12 x = 11
7/25/2019 Solucionario II
74/157
74 Matemtica II
Clave: e
Clave: d
Clave: cClave: d
Clave: a
Clave: e
Clave: c
3. Sea x el nmero de hijos:
50x > 320
x > 6,4 (I)
40x < 320
x < 8 .(II)
De (I) y (II) tenemos
6,4 < x < 8
El nico valor entero es 7
8.13
16
x > 1
Nmeros enteros positivos y menores que 5:
2, 3, 4 (tres nmeros)
4. 5x 1 < 6x + 7
8 < x
x > 8
9.x 1
4+ x 2
5 4
x > 1
6. 3x 1 20
3x 21
x 7
C.S. = [7; +
7. MCM (3; 2; 6) = 6
2(3x 2) + 3(x 3) 5
6x 4 + 3x 9 5
9x 18
x 2
Mayor valor de x es 2
7/25/2019 Solucionario II
75/157
Matemtica II
Clave: d
Clave: b
Clave: b
Clave: d
Clave: c
Clave: a
Clave: c
10.5(x + 1) > 3(x 1)
5x + 5 > 3x 3
2x > 8
x > 4
No puede tomar el valor de 6
11.x: edad de Juan
2x 17 < 35
2x < 52
x < 26 (I)
x2
+ 3 > 15
x2 > 12
x > 24 (II)
De (I) y (II)
24 < x < 26 x = 25
14.2(x 3) + 3(x 2) > 4(x 1)
2x 6 + 3x 6 > 4x 4
5x 12 > 4x 4
x > 8
El menor valor entero x es 9
15.x + 1
2+ x 1
36
MCM (2; 3) = 6
multiplicamos por 6 a cada trmino
6 x + 12
+ 6 x 13
6 6
3(x + 1) + 2(x 1) 36
5x + 1 36
x 7
C.S. = 7; +
12.Sea x la cantidad de patos
x 35 >x
2 2x 70 > x
x > 70 (I)
(x 35) + 3 18 < 22
x 50 < 22
x < 72 (II)
De (I) y (II): 70 < x < 72
Luego: x = 71
16.x 3 < 2x + 2 < x + 5
x 3 < 2x + 2
5 < x
x > 5 ... (I)
2x + 2 < x + 5
x < 3 ... (II)
De (I) y (II): 5 < x < 3
C.S. = 5; 3
13.5x + 3
2+ 2x + 1
4> 1
Multiplicamos por 4 a cada trmino
4 5x + 32
+ 4 2x + 14
> 4 1
10x + 6 + 2x + 1 > 4
12x > 3
x > 14
C.S. = 1/4; +
7/25/2019 Solucionario II
76/157
76 Matemtica II
Clave: c
Clave: b
Clave: b
Clave: d
17.11 6x 1 x < 7 2x
11 6x 1 x
10 5x
2 x
x 2 (I) 1 x < 7 2x
x < 6 (II)
De (I) y (II): 2 x < 6
Valores enteros de x : 2, 3, 4, 5
cuatro
19.Sea x un nmero impar y mltiplo de 3
x x4
< 120
3x4
< 120
x < 160 (I)
20.Sea ab el nmero de 2 cifras:
a + b > 10
b > 10 a (I)
a 2b > 4
a 4 > 2b
a 4
2> b
b 156
De (I) y (II): 156 < x < 160
Luego x impar y mltiplo de 3 es: 159
7/25/2019 Solucionario II
77/157
Matemtica II
Clave: a
Clave: e
Clave: c
21.x + 2
3+ x + 6
5+ x + 3
75
MCM (3; 5; 7) = 105
Multiplicamos a cada trmino por 105
105 x + 2
3 + 105 x + 6
5 + 105 x + 3
7
105 5
35(x + 2) + 21(x + 6) + 15(x + 3) 525
35 x + 70 + 21x + 126 + 15x + 45 525
71x + 241 525
71 x 289
x 4
C.S. = ; 4]
Intervalo no solucin: 4; + 24. x + 14
+ x 12
+ x + 36
10
MCM (4; 2; 6) = 12
Multiplicamos a cada trmino por 12
12 x + 14
+ 12 x 12
+ 12 x + 36
12
3(x + 1) + 6(x 1) + 2(x + 3) 120
3x + 3 + 6x 6 + 2x + 6 120
11x + 3 120
x 11711
... (I)
2x 1
5+ 3x 1
82
MCM (5; 8) = 40
Multiplicamos a cada trmino por 40
40 2x 15
+ 40 3x 18
40 2
8(2x 1) + 5(3x 1) 80
16x 8 + 15x 5 80
31x 13 80
31x 93
x 3 ...(II)
23.2x + 5
3< x 1
2
4x + 10 < 3x 3
x < 13 (I)
22.x 3
2< x < x + 1
3
x 3
2< x
x 3 < 2x
3 < x ...(I)
x 0
x2 5x < 0
x(x 5) < 0
P.C.: x = 0 x 5 = 0
x = 0 x = 5
C.S. = 0; 5
12. 42 x x2> 0
x2+ x 42 < 0
x +7
x 6
(x + 7)(x 6) < 0
P.C.: x + 7 = 0 x 6 = 0
x = 7 x = 6
C.S. = 7; 6
13. x(x 8) + 8 > 4(1 x)
x2 8x + 8 > 4 4x
x2 4x + 4 > 0
(x 2)(x 2) > 0
(x 2)2
> 0x 2
C.S. = {2}
14. x2+ 2x + 1 + x2+ 4x + 4 = x2+ 6x + 9
x2 4 = 0
(x + 2)(x 2) = 0
x + 2 = 0 x = 2
x 2 = 0 x = 2Mayor valor: 2
15. x = 2
2(2)2+ (2m 1)(2) + m 21 = 0
8 + 4m 2 + m 21 = 0
5m = 15
m = 3
16. 40 3x x20
x2+ 3x 40 0
x 5
x + 8
(x 5)(x + 8) 0
P.C.: x 5 = 0 x + 8 = 0
x = 5 x = 8
C.S. = [8; 5]
Clave: c
Clave: d
Clave: a
Clave: d
Clave: e
Clave: e
+
+
+
+
2
0
+
+
0
5
+ +
7 +6
7/25/2019 Solucionario II
81/157
Matemtica II
17. x2 3x 2x
x2 5x 0
x(x 5) 0
P.C.: x