Giải thích các bước giải:
a.Ta có:
$(1-\dfrac{4\sqrt{x}}{x-1}+\dfrac{1}{\sqrt{x}+1}):\dfrac{x-2\sqrt{x}}{x-1}$
$=(1-\dfrac{4\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)})\cdot \dfrac{x-1}{x-2\sqrt{x}}$
$=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)-4\sqrt{x}+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}\cdot \dfrac{x-1}{x-2\sqrt{x}}$
$=\dfrac{x-3\sqrt{x}-2}{x-1}\cdot \dfrac{x-1}{x-2\sqrt{x}}$
$=\dfrac{x-3\sqrt{x}-2}{x-2\sqrt{x}}$
$=\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}$
$=\dfrac{\sqrt{x}-1}{\sqrt{x}}$
b.Ta có:
$(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+3}{x-9}):(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1)$
$=(\dfrac{2\sqrt{x}+\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}):\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}$
$=(\dfrac{3\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}$
$=(\dfrac{3\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)})\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$
$=\dfrac{3\sqrt{x}(\sqrt{x}-3)-3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$
$=\dfrac{-9\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-3)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$
$=\dfrac{-9\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-1)}$
