Giải thích các bước giải:

a.Ta có:

$A=\dfrac{9}{x^3-9x}+\dfrac1{x+3}$ 

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

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

$\to A=\dfrac{9+x^2-3x}{x\left(x+3\right)\left(x-3\right)}$

Khi $x=2\to A=\dfrac{9+2^2-3\cdot 2}{2(2+3)(2-3)}=-\dfrac7{10}$

b.Ta có:

$B=\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}$

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

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

$\to B=\dfrac{3x-9-x^2}{3x\left(x+3\right)}$

$\to C=\dfrac{A}B=\dfrac{9+x^2-3x}{x\left(x+3\right)\left(x-3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}$

$\to C=-\dfrac3{x-3}$