Question

A=(\(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)):(\(\dfrac{3\sqrt{x+1}}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\)) (với x>0,x khác 9)

a,rút gọn biểu thức A

b,tìm x sao cho A>-1

Answer 1

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

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

b: Để A>-1 thì A+1>0

=>\(-3\sqrt{x}+2\sqrt{x}+4>0\)

=>-căn x>-4

=>0<x<16