Cho: A = ( $\frac{x-2}{2x-2}$ + $\frac{3}{2x-2}$ – $\frac{x+3}{2x+2}$ ) : ( 1- $\frac{x-3}{x+1}$ ) ( với x khác ±1) a) Rút gọn biểu thức A. b) Tìm g

Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}):(1-\dfrac{x-3}{x+1})$

$\to A=(\dfrac{x-2}{2(x-1)}+\dfrac{3}{2(x-1)}-\dfrac{x+3}{2(x+1)}):\dfrac{x+1-(x-3)}{x+1}$

$\to A=(\dfrac{x-2+3}{2(x-1)}-\dfrac{x+3}{2(x+1)}):\dfrac{4}{x+1}$

$\to A=(\dfrac{x+1}{2(x-1)}-\dfrac{x+3}{2(x+1)})\cdot\dfrac{x+1}4$

$\to A=(\dfrac{(x+1)(x+1)}{2(x-1)(x+1)}-\dfrac{(x-1)(x+3)}{2(x-1)(x+1)})\cdot\dfrac{x+1}4$

$\to A=\dfrac{(x+1)(x+1)-(x-1)(x+3)}{2(x-1)(x+1)}\cdot\dfrac{x+1}4$

$\to A=\dfrac{x^2+2x+1-x^2-2x+3}{2(x-1)(x+1)}\cdot\dfrac{x+1}4$

$\to A=\dfrac{4}{2(x-1)(x+1)}\cdot\dfrac{x+1}4$

$\to A=\dfrac{1}{2(x-1)}$

b.Để $A=-1002$

$\to \dfrac{1}{2(x-1)}=-1002$

$\to x-1=-\dfrac{1}{2004}$

$\to x=1-\dfrac1{2004}$

$\to x=\dfrac{2003}{2004}$