Đáp án:
Giải thích các bước giải:
$a)_{}$
$ĐKXĐ:_{}$ $x_{}$ $\neq$ $±2,x_{}$ $\neq3$
$Ta_{}$ $có:_{}$
$A=(\dfrac{2+x}{2-x}-$ $\dfrac{2-x}{2+x}-$ $\dfrac{4x^2}{x^2-4}):$ $\dfrac{x-3}{2x-x^2}$
$=[\dfrac{(2+x)^2-(2-x)^2+4x^2}{(2-x)(2+x)}]:$ $\dfrac{x-3}{x(2-x)}$
$=\dfrac{4+4x+x^2-4+4x-x^2+4x^2}{(2-x)(2+x)}.$ $\dfrac{x(2-x)}{x-3}$
$=\dfrac{4x^2+8x}{(2-x)(2+x)}.$ $\dfrac{x(2-x)}{x-3}$
$=\dfrac{4x(x+2)}{(2-x)(2+x)}.$ $\dfrac{x(2-x)}{x-3}$
$=\dfrac{4x^2}{x-3}$
$b)_{}$
$|x-2|=1_{}$ ⇔ \(\left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3(loại)\\x=1\end{array} \right.\)
Tại $x=1_{}$ ta có :
$A=\dfrac{4.1^2}{1-3}$ =$\dfrac{-4}{2}=-2$
$c)_{}$
Để $A>0_{}$ ⇔ $\dfrac{4x^2}{x-3}>0$
⇔$x-3>0_{}$
⇔$x>3_{}$
Vậy để $A>0_{}$ thì $x>3_{}$
