Question

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

a) tìm điều kiện

b) rút gọn

c)tìm x sao cho A<-1

Answer 1

a: ĐKXĐ: x>0; x<>9

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

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

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

\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+2}\)