Giúp mình với còn 1 bài này thôi

Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{x^2-2}{x^2+2x}+\dfrac1{x+2}):\dfrac{x+1}x$

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

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

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

$\to A=\dfrac{x-1}{x+1}$

b.Ta có:

$B=(\dfrac{x+2}{x-2}-\dfrac1{x+2}-\dfrac{x-4}{4-x^2}):\dfrac1{x^2-4}$

$\to B=(\dfrac{x+2}{x-2}-\dfrac1{x+2}+\dfrac{x-4}{x^2-4})\cdot (x^2-4)$

$\to B=(\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-4}{\left(x-2\right)\left(x+2\right)})\cdot (x-2)(x+2)$

$\to B=\dfrac{\left(x+2\right)^2-\left(x-2\right)+x-4}{\left(x-2\right)\left(x+2\right)}\cdot (x-2)(x+2)$

$\to B=\dfrac{x^2+4x+2}{\left(x-2\right)\left(x+2\right)}\cdot (x-2)(x+2)$

$\to B=x^2+4x+2$

c.Ta có:

$C=(\dfrac{x}{x^2-9}+\dfrac1{x+3}+\dfrac2{3-x}):\dfrac3{x+3}$

$\to C=(\dfrac{x}{(x+3)(x-3)}+\dfrac1{x+3}-\dfrac2{x-3})\cdot \dfrac{x+3}3$

$\to C=\dfrac{x+x-3-2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}3$

$\to C=\dfrac{-9}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}3$

$\to C=-\dfrac{3}{x-3}$