Question

rút gọn\(\dfrac{x+\sqrt{X}+10}{X-9}-\dfrac{1}{\sqrt{x}-3}:\dfrac{\sqrt{x}-3}{1}\)

Answer 1

Ta có: \(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}:\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{\left(x+\sqrt{x}+10\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)^2\cdot\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)^2\cdot\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x\sqrt{x}-3x+x-3\sqrt{x}+10\sqrt{x}-30-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)^2}\)

\(=\dfrac{x\sqrt{x}-2\sqrt{x}+6\sqrt{x}-33}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)^2}\)