Question

\(B=\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}-\frac{6}{9-x}\right):\frac{1}{\sqrt{x}-3}\)

Answer 1

Ta có: \(B=\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}-\frac{6}{9-x}\right):\frac{1}{\sqrt{x}-3}\)

\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{\sqrt{x}-3}{1}\)

\(=\frac{x+5\sqrt{x}+6+\sqrt{x}-3+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{1}\)

\(=\frac{x+6\sqrt{x}+9}{\sqrt{x}+3}\)

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