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 giá trị của x để A có giá trị bằng -1002.
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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 $A=-1002$
$\to \dfrac{1}{2(x-1)}=-1002$
$\to x-1=-\dfrac{1}{2004}$
$\to x=1-\dfrac1{2004}$
$\to x=\dfrac{2003}{2004}$