$a)$ $Q=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}+1}-\dfrac{2}{x-1}$ `;(x>0;x!=1)`
$=\dfrac{\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}$
$=\dfrac{\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)-2\left(x-1\right)\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}$
$=\dfrac{x^2-x\sqrt{x}-x+\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}$
$=\dfrac{x^2-x\sqrt{x}-x+\sqrt{x}}{\left(x-1\right)^2}$
$=\dfrac{\left(-1+x\right)\left(x-\sqrt{x}\right)}{\left(x-1\right)^2}$
$=\dfrac{x-\sqrt{x}}{x-1}$
$b)$ Thay `x=9` vào `Q` ta được:
$Q=\dfrac{9-\sqrt{9}}{9-1}$ $=\dfrac{6}{8}$ $=\dfrac{2.3}{2.}$ $=\dfrac{3}{4}$
`⇒` `Q=\frac{3}{4}` khi `x=9`
