Bài 2 :
a. $ĐK:$ $x \neq 1 ; -1 $
B = $\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^{2}}$
= $\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{(1-x)(1+x)}$
= $(x-1)^{2}-(x+1)^{2}+4$
= $x^{2}-2x+1-x^{2}-2x-1+4$
= $-4x+4$
= $-4(x-1)$
= $\frac{-4(x-1)}{(x-1)(x+1)}$
= $\frac{-4}{x+1}$
b. $x^{2}-x=0$
⇔$x.(x-1)=0$
⇔ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Khi x = 0 ta đc :
B = $\frac{-4}{1}$ = $-4$
Khi x = 1 ta đc :
B = $\frac{-4}{2}$ = $-2$
c . B = -3
⇔ $\frac{-4}{x+1}=-3$
⇔ $-4=-3(x+1)$
⇔ $-4=-3x-3$
⇔ $3x=1$
⇔ $x=\frac{1}{3}$
d. B < 0
⇔ $\frac{-4}{x+1}$ < 0
⇔ x+1 > 0
⇔ x > -1
