Question

Thu gọn biểu thức sau :

A=(\(\frac{\sqrt{x}}{\sqrt{x}+3}\)+\(\frac{3}{\sqrt{x}-3}\)).\(\frac{\sqrt{x}+3}{x+9}\)

Answer 1

ĐKXĐ: \(x\ge0;x\ne9\)

\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{x+9}\)

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