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ejercicio 140 libro de baldor resuelto
- 1. 2.-
4.- (x+y)2 - x(x-y)2
y xy
m.c.m.
[x(x+2)2] x(x-y)2
xy xy
= x(x+y)2 - x(x-y)2
xy
= x3+2x2y+xy2-x3+2x2y-xy2
xy
= 4x2y
xy
6._ a + 1+5a = a3-5a+1+5a = a3+1 = (a+1)(a2-a+1)
a2-5 a2-5 a2-5 a2-5
a – a+5 = a2+a-a-5 = a2-5
a+1 a+1 a+1
- 2. (a+1)(a2-a+1) . a2-5 = a2-a+1
a2-5 a+1
N# 8
10._ x3-xy2 = x2+xy
x-y
x3-xy2 = x(x+y)
x-y
x (x2-y2) = x(x+y)
x-y
- 3. x(x+y)(x-y) = x (x+y)
x-y
x (x+y) = x (x+y)
12.-
Ejercicio 14
a2 - a4 x 1-a + 1+ a3 m.c.m: (a-1) (a+1) (a-1) (a+1)=(a2+1)(a2-1)
1-a2 1-a4 a2. a2-1: (a-1) (a+1)
a4-1: (a2-1) (a2+1)=
(a-1) (a+1)(a-1) (a+1)
a2 (a2-1) - (a2)(a2) X 1-a + 1-a3.
(a2+1)(a2-1) a2.
a4- a4-a2. X (1-a)(a2) + a3+1
- 4. (a2+1)(a2-1) a2
a2. X a2-a3 + a3 +1
(a4-1) a2
a2 X a2+1
(a2+1)(a2-1) a2
(a2 ) ( a2+1)
(a2+1)(a2-1) (a2)
R=a2-1
N# 16
Factorización de:
Factorización de:
- 5. Ejercicio 18
1 - 2 + 3 ÷ x + x + 6
x x+2 x+3 x+2 x+3 x2+5x+6
(1)(x+2)(x+3) - [(2)(x)(x+3)] +(3)(x)(x+2) 6 ÷ (x)(x+3) + (x)(x+2) + 6
(x ) ( x+2) ( x+3) (x+2) (x+3)
x2+5x+6 – 2x2-6x-6 + 3x2+6x ÷ x2+3x + x2+2x + 6
(x ) ( x+2) ( x+3) (x+2) (x+3)
- 6. 2x2+5x+6 ÷ 2x2+5x+6
(x ) ( x+2) ( x+3) (x+2) (x+3)
(2x2+5x+6)(x+2) (x+3) m.c.m (x+2)(x+3)
(x ) ( x+2) ( x+3) (2x2+5x+6) x+2 = x+2
x+3= x+3
(x+2) (x+3) (x+2) (x+3) x2+5x+6=
(x+2) (x+3) (x ) ( x+2) ( x+3) (x+2)(x+3)
R= 1
X
20._ 1 x2-36 ÷ x . 1 . 1 _
3 x x2-4 x-36 x-4x
x x
m.c.m
x2-36 = (x+6) (x-6)
x2-4 = (x+2) (x-2)
- 7. 1 (x+6) (x-6) ÷ x . 1 . 1 _
3 x (x+2) (x-2) x-36 x-4x
x x
(x+6) (x-6) . (x+2) (x-2) . x . x__
3x x x2-36 x2-4x
x2 (x+6) (x-6) (x+2) (x-2) = x2 = 1
3x2 (x+6) (x-6) (x+2) (x-2) 3x2 3
22.-
.
.
- 8. 24.-
. a-b
. 1
= —
a-b