Question

\(\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{x-9}\right).\left(\sqrt{x}+\dfrac{\sqrt{x}-9}{\sqrt{x}-1}\right)\)

Answer 1

\(\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{x-9}\right)\cdot\left(\sqrt{x}+\dfrac{\sqrt{x}-9}{\sqrt{x}-1}\right)\left(đk:x\ge0;x\ne9;x\ne1\right)\)

\(=\left[\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\cdot\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\sqrt{x}-9}{\sqrt{x}-1}\right]\)

\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{x-\sqrt{x}+\sqrt{x}-9}{\sqrt{x}-1}\)

\(=\dfrac{3}{x-9}\cdot\dfrac{x-9}{\sqrt{x}-1}\)

\(=\dfrac{3}{\sqrt{x}-1}\)