Question

Cho biểu thức D=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

a) tìm điều kiện

b) rút gọn

c) tìm x sao cho D < -1

Answer 1

a: ĐKXĐ: x>0; x<>9

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

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

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

c: Để D<-1 thì D+1<0

\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}+4< 0\)

\(\Leftrightarrow4-\sqrt{x}< 0\)

hay x>16