QuintoAñoSecundaria

Quinto Año de Secundaria

Solucionario

quinto año de educación secundaria

-1-

CAPÍTULO 2

ANÁLISIS COMBINATORIO Y POTENCIACIÓN (Pág. 34, 35, 36)

Factorial de un número
NIVEL I Resolución 7
1 (n + 3 )!
Resolución 1 · = 10
3 (n + 1)!
E = (n + 2)! – 2(n+1)!
(n + 3)! = 30(n + 1)!
E = (n + 2)(n + 1)! – 2(n + 1)! = (n +1)![n+2–2]
(n + 3)(n + 2)(n + 1)! = 30(n + 1)!
∴ E = n(n + 1)! Rpta.: D
(n + 3)(n + 2) = 30
Resolución 2
∴ n=3 Rpta.: B
7! − 2 × 5! 7 ·6 ·5! − 2·5! 7·6· 5 ! − 2· 5 !
M= = =
6! − 10 × 4! 6·5! − 2·5·4! 6· 5 ! − 2· 5 ! Resolución 8
42 − 2 (x – 1)! + x! + (x + 1)! = 5880
M=
6−2
(x – 1)! + x(x – 1)! + (x + 1)· x ·(x – 1)!= 5880
∴ M = 10 Rpta.: E
(x – 1)![1 + x + (x + 1)·x] = 5880
Resolución 3 (x – 1)!(x2 + 2x + 1) = 5880
1 1 1 1 (x – 1)!(x + 1)2 =5! · 72
E= = = =
4!+ 3! 4· 3!+ 3! 3!(4 + 1) 3!· 5
x–1=5
4 4
E= = Rpta.: E ∴ x=6 Rpta.: B
3!· 4 · 5 5!
Resolución 9
Resolución 4
1 1 (n + 1) 1 (x − 1)! (x + 2 ) = 5
E= − = − x! 3
n! (n + 1)! n!(n + 1) (n + 1)!
n +1 1 n + 1− 1
3(x – 1)!(x + 2) = 5x · (x – 1)!
E= − =
(n + 1)! (n + 1)! (n + 1)! 3x + 6 = 5x
E=
n ∴ x=3 Rpta.: B
∴ (n + 1)!
Rpta.: D

Resolución 10
Resolución 5
(n + 1)!− n! = (n + 1)n!− n! = n![n + 1− 1] m!(n + 1)! m!(n + 1) n!
R= E= =
(n − 1)! (n − 1)! (n − 1)! (m + 1)! n! (m + 1)m! n!

n!n n!· n · n n! n2 n+1
R= = = ∴ E= Rpta.: B
(n − 1)! n(n − 1)! n! m+1

∴ R = n2 Rpta.: B Resolución 11

Resolución 6 11!+10!+ 9! 11· 10· 9· 8!+10· 9· 8!+ 9· 8!
R= =
(n + 2)! = 6 (n + 2)(n + 1)n! = 6 121· 8! 121· 8!
à
n! n! 11 10· 9 + 10· 9 + 9
·
R=
(n + 1)(n + 2) = 6 121
Resolviendo:
∴ R=9 Rpta.: B
∴ n=1 Rpta.: A

-2-


Resolución 12 Resolución 3
 (n + 1)!  (n + 3 )! (n + 2)! − n n + 3 + (n − 2)!
2 − =6 P= ( )
 n!  (n + 2)!
n! (n − 3)!
(n + 2)(n + 1)n! − n n + 3 + (n − 2)(n − 3)!
2· (n + 1)n! (n + 3)(n + 2)! = 6 P= ( )
− n! (n − 3)!
n! (n + 2)!
P = n2 + n + 2n + 2 − n2 − 3n + n − 2
2n + 2 – n – 3 = 6
∴ P=n Rpta.: C
∴ n=7 Rpta.: C

Resolución 4
Resolución 13
( x − 5)! 2 ( x − 4 )!
(x + 6 )! − (x + 2)! = 44 =
( x − 3)! ( x − 2)!
( x + 4)! x!
( x − 5)! 2 ( x − 4 )!
( x + 6 )(x + 5)( x + 4)! − (x + 2)( x + 1) x! = 44 =
( x − 3 ) ( x − 4 ) ( x − 5)! (x − 2) ( x − 3 ) (x − 4 )!
(x + 4 )! x!
1 2
(x + 6)(x + 5) – (x + 2)(x + 1) = 44 = x–2 = 2x – 8
x−4 x−2
8x + 28 = 44
∴ x=6 Rpta.: D
∴ x=2 Rpta.: D
Resolución 5
Resolución 14
( x − 2)!+ (x − 1)! = 720
(n + 1)! (n – 1)! = 36n + (n!)2 x
(n + 1)n(n–1)!(n–1)! = 36n+[n(n–1)!]2 (x–2)! + (x–1)(x–2)! = 720x
(n + 1)n[(n–1)!]2 = 36n + n2[(n–1)!]2 (x–2)!(1+x–1) = 720 x
[(n–1)!]2 [n2 + n – n2] = 36n (x–2)! = 6! x–2= 6
[(n–1)!]2[n] = 36n ∴ x=8 Rpta.: B
(n–1)! = 6
(n–1)! = 3! Resolución 6
(n – 1) = 3 (n + 4)! − (n + 3)! = 25
∴ n=4 Rpta.: C (n + 2)! (n + 2)!

(n + 4 )(n + 3 )(n + 2 )! − (n + 3 )(n + 2 )! = 25
NIVEL II (n + 2 ) (n + 2 )!
n2 + 3n + 4n + 12 – n – 3 = 25
Resolución 1
n2 + 6n + 9 = 25
n!
R= − n2 ∴ n=2 Rpta.: C
(n − 2)!
n (n − 1)(n − 2)! Resolución 7
R= − n2 = n2 − n − n2
(n − 2)! 
A=
(n + 1)!+ n!   (2n + 3)! 
 
∴ R = –n Rpta.: D 
 (2n + 1)!+ (2n + 2)!   (n + 2)! 
 

Resolución 2 
A=
(n + 1)n!+ n!  (2n + 3)(2n + 2)(2n + 1)!
·
(2n + 1)!+ (2n + 2)(2n + 1)!
  (n + 2)(n + 1· n!
)
n (n + 1)!− n! n (n + 1) n!− n! n· n!(n + 1− 1)
M=  =  =
(n − 1)! (n − 1)! (n − 1)! =
n!  n + 2 
  ·
( 2n + 3 )(2n + 2) (2n + 1)!
(2n + 1)!  2n + 3 
  ( n + 2 )(n + 1)n!
n· n· n! n· n· n (n − 1)!
M= =
(n − 1)! (n − 1)! 2 (n + 1)
=
n +1
∴ M = n3 Rpta.: C
∴ A=2 Rpta.: B

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Resolución 8 (13· 12)2 (11!)2 13· 12· 11 10!
·
−
(n + 7 )! ⋅ (n + 5 )! = 10! 2
(12 + 1) (11!) 2 10! (1+ 11)
(n + 6 )!+ (n + 5 )!
(13· 12)2 − 13· 12· 11
(n + 7)!(n + 5)! = 10! (13 )2 12
(n + 6) · (n + 5)!+ (n + 5)!
(12)2 – 13· 11
(n + 7 )! (n + 5)! = 10!
(n + 5)! [n + 6 + 1] ∴ 1 Rpta.: A
Resolución 12
(n + 7)(n + 6)! = 10! (n + 6)! = 10! (119!)x!! (5!)x!! = (5!!23!)24
(n + 7 )
(119! 5!)x!! = (5!!)23!· 24
n + 6 = 10
(119! 120)x!! =(5!!)24!
∴ n=4 Rpta.: E
(120!)x!! = (5!!)24!
Resolución 9 (5!!)x!! = (5!!)24!

R=
(a!!+ 2)!− 2(a!!+ 1)! = (a!!+ 2)(a!!+1)!− 2(a!!+ 1)! x!! = 24! x!! = 4!!
(a!!+ 1)! (a!!+ 1)! ∴ x=4 Rpta.: B

R=
(a!!+ 1)! (a!!+ 2 − 2) Resolución 13
(a!!+ 1)!
5 5 5
∴ R = a!! Rpta.: B = =
5!+ 4!+ 3! 5· 4· 3!+ 4· 3!+ 3! 3!(20 + 4 + 1)
Resolución 10 5 1 4 4
= = = Rpta.: D
E = (n!! – 1)!(n!–1)!(n–1)!n–n!!! 3!· 25 3· 2· 1 5 5· 4· 3· 2· 1 5!
·
E = (n!!–1)!(n!–1)!n! – n!!!
E = (n!!–1)! n!! – n!!! Resolución 14
E = n!!! – n!!! (n + 2)! = 5+
(n + 12)!
∴ E=0 Rpta.: C n! (11+ n)!
Resolución 11 (n + 2)(n + 1)n! = 5 + (n + 12)(n + 11)!
(13!)2 13! n! (n + 11)!
2
−
2 10!+ 11!
(12!) + 2 (12!11!) + (11!) (n+2)(n+1) = 5+n+12
(13!)2 − 13! n2 + 3n+2 = 5+n+ 12
(12!+ 11!)2 10!+ 11! n2 + 2n = 15
∴ n=3
(13· 12· 11!)2 − 13· 12· 11· 10!
(12· 11!+ 11!)2 10!+ 11· 10! ∴ Suma valores = 3 Rpta.: C

ANÁLISIS COMBINATORIO (Pág. 45, 46)

NIVEL I
Resolución 2
Resolución 1

5 pantalones 3 blusas

N° maneras = 5 × 3
N° maneras = 6 × 4
∴ N° maneras = 15 Rpta.: C
∴ N° maneras = 24 Rpta.: D

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m 5
...................← Personas
V2 = 20
5
--------------- ← asientos
m! m (m − 1) (m − 2 )! N° maneras = 5· 4· 3· 2· 1
= 20 = 20
( m − 2 )! (m − 2)! ∴ N° maneras = 120 Rpta.: C
m(m–1) = 4 × 5

∴ m=5 Rpta.: C Resolución 10

Resolución 4
A B C D ← asientos
N° maneras = 6 · 5 · 4 · 3
Una persona debe estar fija y las otras 4 las permuta-
∴ N° maneras = 360 Rpta.: B
mos.
N° maneras = 4!
Resolución 5
∴ N° maneras = 24 Rpta.: B
3 : anillos:
4 : dedos
N° maneras = 4· 3· 2 Resolución 11

∴ N° maneras = 24 Rpta.: C N = a b c d > 6000
6523

Resolución 6 N° maneras = 1· 3· 2· 1

10 : amigas ∴ N° maneras = 6 Rpta.: D

6 : invitadas Resolución 12
10· 9· 8· 7
N° maneras = C10
6 = 8· 7· 6· 5
1 2· 3· 4
· C8 =
4
1 2· 3· 4
·
∴ N° maneras = 210 Rpta.: B ∴ N° cuadriláteros = 70 Rpta.: B

n N = abc
  = 15
 4 números: {1; 2; 3; 4; 5}
n (n − 1)(n − 2 )(n − 3 ) N° maneras = 5· 4· 3
= 15 ∴ N° maneras = 60 Rpta.: D
1 2· 3· 4
·
n(n–1)(n–2)(n–3) = 6· 5· 4· 3
Resolución 14
∴ n=6 Rpta.: B  n + 1  n 
 :  ... (1)
 n   n − 1
Resolución 8
Entonces:
x x
C5 + C6 = 28  n + 1  n + 1   n + 1
 = =  = n+1
C5 + C6 = C6 +1 = 28
x x x  n   n + 1− n   1 

( x + 1) x (x − 1)( x − 2)( x − 3)( x − 4) = 28  n   n  n
1 2· 3· 4· 5· 6
·  = = =n
 n − 1  n − (n − 1)   1 
(x+1)x(x–1)(x–2)(x–3)(x–4) = 8·7·6·5·4·3 En (1):

∴ x=7 Rpta.: C n+1
∴ Rpta.: D
n

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x
C5 = 21  p + q (p + q)! = (p + q)!
 =
 p  p! (p + q) − p  !
  p! q!
x (x − 1)(x − 2)(x − 3 )(x − 4 )
= 21
1 2·3· 4· 5
· Además:
x(x–1)(x–2)(x–3)(x–4) = 7· 6· 5· 4· 3  p + q  p + q   p + q
 = = 
∴ x=7 Rpta.: E  q   (p + q) − q   p 

∴ Son equivalentes I y II Rpta.: B
NIVEL II

4 : biólogos → se escogen 2

3 : químicos → se escogen 2

De ida: 2 + 2·3 + 1= 9 caminos 5 : matemáticos → se escogen 3
De venida: 2 + 2· 3 + 1 = 9 caminos
N° maneras = 9· 9 = 81 N° maneras = C4 · C3 · C5
2 2 3
Quitamos los 9 caminos de ida.
N° maneras = 81 – 9
 4·3  3·2  5·4·3 
∴ N° maneras = 72 Rpta.: B N° maneras =  1· 2  ·  1· 2  ·  1· 2 · 3 
     

Resolución 2 ∴ N° maneras = 180 Rpta.: C
N° maneras = 7· 6 · 5
∴ N° maneras = 210 Rpta.: D Resolución 9

x
Resolución 3   = 0 ..... (1)
 10 
Números = {1; 2; 3; 4; 5; 6; 7; 8; 9} Se sabe que:
N = a bc d e  m
↓ ↓↓ ↓ ↓  =0 ⇔ m<n ∧ m>0
98765 n
N° formas = 9· 8·7· 6· 5 En (1): x < 10 x = {1; 2; 3; 4; 5; 6; 7; 8; 9}
∴ N° formas = 15120 Rpta.: C
Producto = 1· 2· 3· 4· 5· 6· 7· 8· 9
Resolución 4 ∴ Producto: 9! Rpta.: D
L I B R O → 5 letras
N° palabras = 5!
Resolución 10
∴ N° palabras = 120 Rpta.: B
 n + 1  n   n   n + 1
−
Q= + + + 
Resolución 5  2   1  n − 1  n − 1
25· 24 Se sabe que:
C25 =
2
12
·
 m  m 
 = 
∴ N° partidos = 300 Rpta.: D  n   m − n

 n + 1  m + 1
n  n   n
Resolución 6  =  y  = 
 n − 1  2   n − 1  1 

N° diagonales = C8 − N° lados Luego:
2
 n + 1  n    (n + 1)n 
8 ·7 Q = 2   +   = 2  + n
N° diagonales = 1· 2 − 8   2   1   12
· 

∴ N° diagonales = 20 Rpta.: B ∴ Q = n2 + 3n Rpta.: B

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 n  n −1 
 +  = 99
 n + 1  n − 2 
Se sabe que:
 m
 =0 ⇔ m < k N° maneras = 1· 5· 4· 3· 2· 1
k 
∴ N° maneras = 120 Rpta.: E
 n 
 =0
 n + 1 Resolución 13
Luego: 3 : entradas → se toma 1
n −1  n −1  3 : de fondo → se toma 1
0+  = 99   = 99
 n − 2  (n − 1) − (n − 2)  5 : postres → se toma 1
3 3 5
N° maneras = C1 · C1 · C1
 n − 1
  = 99 n – 1 = 99
 1  N° maneras = 3· 3· 5
∴ n = 100 Rpta.: D ∴ N° maneras = 45 Rpta.: A

BINOMIO DE NEWTON (Pág. 51, 52, 53)

NIVEL I

Resolución 1

A) (x–2y)5 = x5 – 5x4 · 2y + 10x3· (2y)2 – 10x2 · (2y)3 + 5x(2y)4 – (2y)5

= x5 – 10x4y + 40x3y2 – 80x2y3 + 80xy4 – 32y5

B) (1 + 3a)7 = 17 + 7(1)6(3a) + 21(1)5(3a)2 + 35(1)4(3a)3 + 35(1)3(3a)4 +21(1)2(3a)5 +

7(1)(3a)6 + (3a)7

=1 + 21a + 189a2 + 945a3 + 2835a4 + 5103a5 + 5103a6 + 2187a7

C) (1–b)11 = 111 – 11(1)10(b)1 + 55(1)9b2 – 165(1)8b3 + 330(1)7b 4 – 462(1)6b5 +

462(1)5b6 – 330(1)4b7 + 165(1)3· b8 – 55(1)2·b9 + 11(1)b10 – b11

= 1 – 11b + 55b2 – 165b3 + 330b4 – 462b5 + 462b6 – 330b7 + 165b8 – 55b9

+ 11b10 – b11

6
 1 6 5 -1 4 -1 2 3 -1 3 2 -1 4 -1 5 -1 6
D)  x − x  = x – 6(x) ·(x ) + 15(x) (x ) –20(x) (x ) + 15(x) (x ) – 6(x)(x ) + (x )
 
= x6 – 6x4 + 15x2 – 20 + 15x-2 – 6x-4 + x-6
4
 2 1
E)  z + 2  = (z2)4 + 4(z2)3(z-2) + 6(z2)2(z-2)2 + 4(z2)(z-2)3 + (z-2)4
 z 

=z8 + 4z4 + 6 + 4z-4 + z-8
6
 3 x3 
F)  4−  = (3x-4)6 – 6(3x-4)5(4-1x3) + 15(3x-4)4(4-1x3)2 – 20(3x-4)3(4-1x3)3 +
x 4 
 

15(3x-4)2(4-1x3)4 – 6(3x-4)(4-1x3)5 + (4-1x3)6

−24 729 −17 1215 −10 135 −3 135 4 9 11 1 18
= 729x − x + x − x + x − x + x
2 16 16 256 512 4096

-7-

 11 A) (2x – y)4
A) (x – y)11 ; t7 = t6+1 =   x11−6 y6
6  4 1
coef(t2) = coef(t1+1) =  1  2 (−1)
3
∴ t7 = 462x5y6  
∴ coef(t2) = – 32
 21
B) (a + b)21 ; t5 = t4+1 =   a21− 4b4 B) (3a + b)6
4
∴ t5 = 5985 a17 b4  6 4 2
coef(t3) =   (3 ) (4 ) = 19440
10 10 −9 9
 2
 1 1  10   1   −1
C) a − b ; t10 = t9+1 =    b
   9  a     x 2 y2 
10

C)  − 
 y x 
∴ t10 = – 10a-1 b-9  
10  10−8
8
 2 2 
7
 7  2 7 − 7  −2 
7
coef(t9) = coef(t8+1)=  10  (1) ( −1)8 = 45
D) x y − 2 

 xy 

; t8 = t7+1 =  7  x y
 
( ) 2
 xy 
 
8
 

D) (–a + 12)5
∴ t8 = –128x-7y-14  5 5−4
coef(t5) = coef(t4+1) =  4( −1) (12)4 = −5·124
 10   
10−10
E) (2a – b)10 ; t11 = t10+1 =  10  (2a ) ( −b )10
  E) (p 2 v 2 –1)14
∴ t11 = b10
 14 
coef(t8) = coef(t7+1) =  7  (1)14-7(–1)7 = –3432
4 1  
 1   4  4 −1  −1 
F)  1−  ; t2 = t1+1 =   (1)  
 xyz  1   xyz  F) (2x2y + xy3)8
8
∴ t2 = –4x-1y-1z-1 coef(t5) =   (2)8-4 (1)4 = 1120
 4

Resolución 4
5
 2 1
( )
5
2 −1
 3x − x  = 3x − x
 
5
 2 1 2 5 2 4 -1 2 3 -1 2 2 2 -1 3 2 -1 4 -1 5
 3x − x  = (3x ) – 5(3x ) (x ) + 10(3x ) (x ) – 10(3x ) (x ) + 5(3x )(x ) – (x )
 
5
 2 1 10 7 4 -2 -5
 3x − x  = 243x – 405x + 270x – 90x + 15x – x
 
A) coef(t4) = –90
B) t3
C) No existe el término independiente de x:

Resolución 5
Nos piden:
 2
12 (x)3k-24 = x-3 3k – 24 = –3 k=7
3
 2 − 3xy 
x y



(
= 2x y −2 −1
− 3xy )
3 12
Luego:
tk+1 = t7+1 = t8
12−k
 12   2 
(−3xy3 )
k
A) tk+1 =  k   2   12 
  x y 
  B) tk+1 =  k  (2)12-k (–3)k (x)3k-24 (y)4k-12
 
 12  (y)4k-12 = y12 4k – 12 = 12 k=6
tk+1 =  k  (2)12-k (–3)k (x)3k-24 (y)4k-12
 
∴ tk+1 = t6+1 = t7

-8-


 12   3 3 −k
C) tk+1 =  k  (2)12-k (–3)k (x)3k-24 (y)4k-12 =  k  (3 ) (−1)k (q)15−6k
   
(x)3k-24 = x0 3k – 24 = 0 k=8 (q)15-6k = q9 15 – 6k = 9 k=1
∴ tk+1 = t8+1 = t9  3  3 −1 1 15−6·1
t1+1 =   (3 ) ( −1) (q)
 12  1 
D) tk+1 =  k  (2)12-k (–3)k (x)3k-24 (y)4k-12
  ∴ t2 = –27q9
y4k-12 = y0 4k – 12 = 0 k=3
Resolución 7
∴ tk+1 = t3+1 = t4
( )
10
2x+ 3
Resolución 6
10 
(2p + q)11
( ) ( ) ( ) ( 3)
A) 10 10 9
2 x+ 3 = 2x +  2x
 11  1
tk+1 =  k  (2p)11-k(q)k
   10 
( ) ( 3)
8 2
+  2x
qk = q9 k=9 2
 11
( )
10
tk+1 = t9+1 = t10 =  9  (2p)11-9(q)9 ∴ 2 x+ 3 = 32x10 + 160 6 x9 + 2160x8 + ...
 
∴ t10 = 220 p2q9
Resolución 8
10
 1
B) q−  (1 + 3x2)6
 pq 
6
(3x2 )  6 k
6 −k k 2k
 10   −1 
k tk+1 =  k  (1) =   (3 ) ( x )
tk+1 =   q10−k     k 
k  pq 
6 0 2·0
 10  k −k 10− 2k t0+1 =  0  (3 ) (x ) t1 = 1
=   (−1) (p ) (q)  
k 
(q)10-2k = q9 10 – 2k = 9  6 6 2·6
t6+1 =  6  (3) ( x ) t7 = 729x12
 
1
k=
2 Luego:
t1 · t7 = 1· 729x12
Como k ∈
∴ Producto de los coeficientes = 729
∴ No existe el término
C) (p2 – q3)7
NIVEL II
 7 2
( ) ( )
7 −k k
tk+1 =  k  p −q3
  Resolución 1

 7 k 14 −2k (x – 3y)5
=   ( −1) (p ) (q)3k
k 
 5 5− 5
(q)3k = q9 3k = 9 k=3 t6 = t5+1 =  5  (x ) (−3y )5
 
Luego:
∴ t6 = – 243y5 Rpta.: D
 7 3 14− 2·3
t3+1 =  3  (−1) (p) (q)3·3
 
∴ t4 = –35p8 q9 Resolución 2

3 (2 – x)11
 5 1
D)  3q −   11 11− 7
 q
t8 = t7+1 =  7  (2) ( −x )7
 
k
 3
( )
3 −k  −1
tk+1 =   3q
5 t8 = –5280x7
 
k   q
∴ Coeficiente = – 5280 Rpta.: D

-9-

n
(2a + b)5 2 x
 + 
 x 2
 5 5−1 1
t2 = t1+1 =   (2a ) (b ) = 80 a b
4 n−k K
 n  2  x  n  n− 2k
 1 tk+1 =      2  =  k  ( 2) ( x )2k −n
k x     
∴ Coeficiente = 80 Rpta.: C Para el término independiente:
(x)2k-n = x0 2k – n = 0 n = 2k
Resolución 4 Pero: k + 1 = 4 k=3
7 Entonces: n = 2· 3 n=6
 y
 3x −  Luego: tk+2 = t3+2 = t5
 2
La expresión tiene 7 + 1 = 8 términos  6  6−8 8−6 15 2
t5 =   ( 2 ) (x ) = x
∴ No hay término central Rpta.: E  4 4

15
Resolución 5 ∴ coef(t5) = Rpta.: C
4
(2x – y)6
6 6− 3 Resolución 9
t4 = t3+1 =   (2x ) ( −y )3
 3
13
 3 x2 1 
∴ t4 = –160x3 y3 Rpta.: D  +5 
 2 
x

Resolución 6

( ) (x )
13 − k k

tk + 1 = ( 13 ) 2 −1 x 3
4 2 1
−
 1 5
x − 2 
k

 x 
26 13k
k  13  k −13 −
 4  4 −k  −1   4  k 4 −3k tk+1 =   ( 2) ( x ) 3 15
tk+1 =   ( x )  2  =   ( −1) ( x )
k   x  k  k

Del dato: El término indenpendiente:
26 13k
4 − 0 26 13k
x4-3k = x0 4 – 3k = 0 k=
3 (x ) 3 15 =x − =0 k = 10
3 15
Como k ∈ tk+1 = t10+1 = t11
∴ No hay término independiente Nos piden el t10 k=9
26 13·9
Rpta.: E  13  9−13 −
t10 =   ( 2) (x ) 3 15
Resolución 7 9 
(2x – 1)5 13
715 15
 5 5 −k k  5 5 −k k 5 −k ∴ t10 = x Rpta.: A
tk+1 =   (2x ) ( −1) =   (2 ) ( −1) ( x ) 16
k  k 
Resolución 10
 5  5− 2 2 5− 2
t3 = t2+1 =   (2 ) ( −1) ( x ) = 80x 3 120
2  1
 x + 
 x
 5  5−4
t5 = t4 + 1 =   (2 ) ( −1)4 (x )5−4 = 10x  120  120−k  1   120  120−2k
k
 4
tk+1 =   (x )  x  =  k x
 k     
t3
= 72 80x 3
Luego: = 72
t5 10x Como es de grado 100
∴ Rpta.: C 120 – 2k = 100 k = 10
x = ±3
∴ tk+1 = t10+1 = t11 Rpta.: E

- 10 -




9 (1 + x)3n
2 0,5 
 0,4x + x 
   3n  3n−k k  3n  k
tk+1 =   (1) (x ) =   x
9 −k  0,5 k
k k
9
(
tk+1=   0,4x
k 
2
)  x 
   3n  k +1
tk+2 =  x
 9  9− 2k k − 9 18− 3k  k + 1
=   (2 ) (5 ) ( x )
k   3n  2k −4
t2k-3 =  x
Término independiente:  2k − 4 

(x)18-3k = x0 18 – 3k = 0 k=6 Como los coeficientes son iguales se tiene:

 9  9 −2·6 6 −9 18 −3·6  3n   3n 
Luego: t6+1 =   ( 2) ( 5) ( x )  =  (k + 1) + (2k – 4) = 3n
 6  k + 1  2k − 4 
3k – 3 = 3n
t7 = 0,084 Rpta.: C
∴ k = n+1 Rpta.: A

BINOMIO DE NEWTON CON EXPONENTE NEGATIVO Y/O FRACCIONARIO
Pág. 58

Resolución 5  1   1   1 
  2   2 − 1  2 − 2  
1
t4 =       32x =  1  ⋅ 32x
(1− 2x ) 5  1· 2· 3   16 
 
 
 1 1  
t5  1
= T4 +1 =  5  (1)5 −4 ( −2x )4 = 5  4
4
   4  16x ∴ t4 = 2x

  1  1  1  1  Resolución 7
  5  5 − 1 5 − 2  5 − 3  
       · 16x 4
 1 2· 3· 4
·   −3 
  E= 
   33 

 −21  4 E=
( −3)( −3 − 1)( −3 − 2)( −3 − 3).....( −3 − 32)
= 16x 1 2· 3· 4· 5· ..... ·33
·
 625 

E=
(−3 )( −4 )( −5 )(−6 ) · ..... · ( −35 ) = (−1)31 ( −34 )( −35 )
−336 4 1 2· 3· 4· 5 · ..... · 33
· 12
·
∴ t5 = x
625
∴ E = –595


1  −15   −15   −15 
 1 2 E= + + 
 1 + x3   3   4   5 
4 
   −15 + 1  −15 
E= + 
 4   5 
1 3
 1 −3  1   1
 2   1 2  3  2  −14   −15 
  E= + 
t4 = t3+1 =    4 
3   x  =   32x
3   4   5 
     
E=
(−14 )(−15)(−16 )(−17 )
1 2· 3· 4
·

- 11 -

+
( −15)( −16 )( −17 )( −18 )( −19 ) Resolución 10
1 2· 3· 4· 5
· −2
 1 −3 −9 
∴ E = – 9248 2x − x 
 

Resolución 9 k
−2 −k  9
(
 −2 
)  −2  k + 2 6 − 3k
 −  (2 ) ( x ) 2
tk +1 =   2−1x −3
1 x 2  =  k
(x 2
−3 ) 2
 k
Término indenpendiente:
   

 1 1  1 3k
 2  2 −2
t3 = t2+1 =   x
2  ( )
  −3
2 ( −3 )2 = 2 x ( 9 )
 
2 
6−
2
=0 k=4
   
 1  1  Entonces:
  − 1
( )
2 2  −9 3·4
t3 =   9x −3 = 3  −2  4 +2 6−
1· 2 8x t4+1 = t5 =   (2 ) (x ) 2
 4
−9 ( −2)( −3 )( −4 )( −5)
Si: x=3 t3 = (2)6
8· 33 t5 = 1 2· 3· 4
·

∴ t3 = (–24)-1 ∴ t5 = 320

CAPÍTULO 3
LOGARITMACIÓN (Pág. 93, 94, 95, 96)

NIVEL I Resolución 5
5
Resolución 1 log x 2 = 0,4
2
log a = x 5 2 2 2
log x = logx =
log 10a = log10 + loga = 1 + loga 5 5 5

∴ log10a = 1 + x Rpta.: E ∴ logx = 1 Rpta.: B

log p = x log p = q
1 p
log 3 p = logp log   = log p − log r
3 r
x p
∴ log 3 p = Rpta.: D ∴ log   = q − log r Rpta.: B
3 r

loga = m ; logb = n  1
logx + log   = logx + logx-2 = logx – 2logx
 x2 
a 1 a 1
log = log   = (loga − logb )
b 2 b 2  1
∴ logx + log   = –logx Rpta.: C
 x2 
a m−n
∴ log = Rpta.: B
b 2 Resolución 8
2
Resolución 4 2 2
log5 3 25 = log5 5 3 = log5 5 = · 1
3 3
log 103 = 3log10 = 3· 1
2
∴ log 103 = 3 Rpta.: D ∴ log5 3 25 = Rpta.: D
3

- 12 -


Resolución 9 7 4
log 102 + log2 2 − log 5
5
logx–3 = logx – 3log10 = logx– log103
= logx–log1000 2log 10 + 7log 2 − 4log 5
2 5

 x  2+7–4
∴ logx–3 = log   Rpta.: E 5 Rpta.: B
 1000 


log2 a = x log0,01+ log 0,0081= log10-2 + log (0,3)4
0,3 0.3
x + 1 = log2 a + log2 2
= –2log10 + 4 log (0,3)
∴ x + 1 = log2 2a Rpta.: D (0,3)

=–2+4= 2 Rpta.: C
log(a3–b3)= log(a–b)(a2+ab+b2) log 0,25 + log 0,125 − log 0,0625
2 2 2
log(a3–b3) = log(a–b) + log(a2+ab+b2)
log
(0,25)(0,125) = log (0,03125)
Rpta.: D 2 0,0625 2 (0,0625 )

Resolución 12
log (0,5) = log 2−1 = −1log 2
log(x2–x) = logx(x–1) 2 2 2

∴ log(x2–x) = logx + log(x–1) =–1 Rpta.: E
Rpta.: A
Resolución 20
Resolución 13
 1  1  1 
log   − log   + log 
5  125 
11
3 2 11 12 2  16  3  81 
log 216 6 = log 63 = : log6 6
12 3 5
36 5 36
65 log 2−4 − log 3−4 + log 5−3
2 3 5
55
= Rpta.: C −4log 2 + 4log 3 − 3log 5
36 2 3 5

–4+4–3
Resolución 14
∴ –3 Rpta.: D
log 0,064 = x log (0,4)3 = x
0,4 0,4
Resolución 21
3log 0,4 = x
0,4 log3 = 0,47 , log5 = 0,70

∴ x=3 Rpta.: D log75 – log125 + log45 =
75 · 45
log = log27 = log33 = 3log3
Resolución 15 125
−2 = 3(0,47)
2 9
log x = −2 x=  x=
2 3 4 =1,41 Rpta.: B
3
Rpta.: E
Resolución 22
Resolución 16
log2 = 0,30 ∧ log5 = 0,70
2
log (a2 − 2ab + b2 ) = log (a − b) 35
(a −b) (a −b) log35 – log14 = log
14
= 2 log (a − b) = 2 Rpta.: E
(a −b)
5
= log = log5 − log2
2
Resolución 17
log 35 – log14 = 0,70 – 0,30
log 100 + log 128 − log 625
2 5 ∴ log35 – log14 = 0,40 Rpta.: B

- 13 -

log 2 log 2
1
+
1
log 36 log 36
= log (2) + log (3)
36 36
243
3
= 35 ( ) 3

2 3
log 2 5·log 2 log 25
3 3 3
= log (2· 3) = log 6 243 = (3) = (3) = 25
36 36
log 2
1 ∴ 243 3 Rpta.: E
= 32
1 2
= log (36) = log 36
36 2 36
Resolución 29
1
= Rpta.: C logx + log(x–3) = 1
2
logx (x–3) = log10
Resolución 24 x(x–3) = 10

log 3 = x ∴ x=5 Rpta.: C
2

log 64 = log 26 = 6log 2
24 24 24 Resolución 30
 1   1  log x log3
= 6  = 6  10 − 10 = 2x − 5
log 24 
 2 log (8· 3) 
 2
x – 3 = 2x – 5
 1   1  ∴ x=2 Rpta.: B
= 6  = 6 
log 8 + log 3 
 2  3log 2 + x 
2 2 NIVEL II

6
= Rpta.: B Resolución 1
3+x
3x
 1 1
( )
x −4 2
log  =x = 2 2 2 =2
Resolución 25 2 2  16  16

3
log (5x − 3) − log x = 1 −4 = x
2 2 2

(5x − 3) 8
log = log 2 ∴ x=− Rpta.: A
2 x 2 3

5x − 3 Resolución 2
=2 5x – 3 = 2x
x
∴ x=1 Rpta.: B (I) log 32 = 5 32 = 25 ... (V)
2

Resolución 26 (II) log 1= 0 1=(2000)0 .... (V)
2000

log (2x + 21) − log x = 2  1 1
3 3 (III) log   = −4 = 2−4 ... (V)
2  16  16
 2x + 21 2 2x + 21 2
log   = log3 3 =3 ∴ VVV Rpta.: D
3 x  x

2x + 21 = 9x Resolución 3
∴ x=3 Rpta.: A log 27 = a
12
Resolución 27
log2 24 4log2 2
log a + logb = log(a + b) log 16 = =
6 log2 2 + log2 3 log2 2 + log2 3
log a · b = log(a+b) a·b = a + b a(b–1) = b
4 ........................... (1)
b log 16 =
∴ a= Rpta.: D 6 1 + log2 3
b −1

- 14 -


3log2 3
Pero: log 27 = a =a 52  5
12 2log2 2 + log2 3 = log = 2log  
2 22 2 2

3log2 3 2a
=a log2 3 =   10  
2 + log2 3 3−a = 2 log  2   = 2[log 10 − 2log 2]
2
 2  2  2

Reemplazando en (1) 1 
= 2 log 10 − 2 = 2  − 2 
 
4  10x   x 
log 16 =
6 2a 2 − 4x
1+ =
3−a Rpta.: D
x
12 − 4a
∴ log 16 = Rpta.: E Resolución 8
6 3+a
log y
Resolución 4 (log5 x ) 5 =y
log 3 · log 4 · log 5 · log 6 ..... log 1024 log y
log (log x )
2 3 4 5 1023 5 
= logy

 5 

log3 log4 log5
log1024 log6
· · · ..... log y log (log x ) = logy
log2 log3 log4 log5 log1023 5  5 

log (log x ) = log5 log x = 5
log1024 log210 5 5
= = 10 Rpta.: B
log2 log2 x= 55 = 3125
∴ ∑ cifras = 11 Rpta.: C
Resolución 5
log2 = a ∧ log3 = b Resolución 9
log 4 + log 1 4
3
log 752 =
2
3
2
log75 = log 52 · 3
3
( ) E=
2
2 =
log2 22 + log −1 22
2

=
2 2
log52 + log3 = [2log5 + log3]
log 243 + log 1 81 log 35 + log −1 34
3 3
3 3
( )
3  3

2  10   2−2
= E=
 2log   + log3 5−4
3  2 
2 ∴ E=0 Rpta.: E
=  2 (log10 − log2) + log3
3 
Resolución 10
2
=  2 (1− a ) + b
3 
log 2 = a ∧ log 3 = 2b
5 5
2
∴ log 3 752 =
3
[b − 2a + 2] Rpta.: D log
5
1
2 5
1
300 = log 300 = log 102·3 
2 5
 
 ( )
1 1
Resolución 6 = log 10 + log 3 = [2log 10 + log 3]
2
2 5 5  2 5 5
2 4 6
logx = + − 1
7+ 5 11 + 7 11 + 5 =  2 (log 5 + log 2 ) + log 3 
2 5 5 5 
0
=
( 7+ 5 )( 11 + 7 )( 11 + 5 ) =
1
2
 2 (1 + a ) + 2b 

logx = 0 x = 100
∴ x=1 Rpta.: B ∴ log 300 = a + b + 1 Rpta.: E
5

Resolución 7
Resolución 11
log2 = x 2 = 10x
log(2–x) + log(3–x) = log2 + 1
 2,5 
log 2,5 − log (0,4) = log  log(2–x) + log(3–x) – log2 = log10
2 2 2  0,4 


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