Bài 2:
a) A= [(4x+5)/(2x+2) + 4x/(x^2-1) - (7+2x)/2(x+1)] : 1/(x^2-1)
A=[(4x+5)/2(x+1) - (7+2x)/2(x+1)+ 4x/(x^2-1) ] : 1/(x^2-1)
A= [(4x+5-7-2x)/2(x+1) + 4x/(x-1)(x+1)] : 1/(x^2 - 1)
A=  [2(x-1)/2(x+1) + 4x/(x-1)(x+1)] : 1/(x^2 - 1)
A=  [(x-1)^2/(x+1)(x-1) + 4x/(x-1)(x+1)] : 1/(x^2 - 1)
A=[(x^2 - 2x +1 + 4x)/(x+1)(x-1)] : 1/(x-1)(x+1)
A= (x+1)^2/(x+1)(x-1) : 1/(x-1)(x+1)
A= (x+1)/(x-1) . (x-1)(x+1)/1
= (x+1)^2(x-1)/(x-1) = (x+1)^2
b) Để A =1,
A = (x+1)^2 = 1
=> x+1 = 1 hoặc x+1=-1
=> x = 0 hoặc x = -2
Bài 3:
a) x^3 + 6x^2 + 12x + 8 = 0
(x+2)^3 =0
=> x + 2 =0
=> x =-2
b) x^2 - 2x - 3 =0
(x^2 + x) - (3x+3)=0
x(x+1) - 3(x+1)=0
(x-3)(x+1)=0
=> x=3 hoặc x =-1
c) 5x^2 - 4(x^2 - 2x+1) - 5 = 0
5x^2 - 5 - 4(x-1)^2 =0
5(x-1)(x+1) - 4(x-1)^2=0
(x-1)(5x + 5 - 4x + 4)=0
(x-1)(x+9)=0
=> x=1 hoặc x=-9