Question

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+6\sqrt{x}+9}{9-x}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\) với x≥0;x≠9
rút gọn

Answer 1

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+6\sqrt{x}+9}{9-x}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\left(dkxd:x\ge0,x\ne9\right)\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)-\left(x+6\sqrt{x}+9\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-6\sqrt{x}-x-6\sqrt{x}-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-9\sqrt{x}-9}{x-9}\) với \(x\ge0,x\ne9\)