Question

Cho biểu thức A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{1}{2}\right)\) (Với x\(\ge0,x\ne9\))

a.Rút gọn A

b.Tìm x để A<\(-\dfrac{1}{2}\)

Answer 1

a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{-6}{\sqrt{x}+3}\)

b: Để A<-1/2 thì A+1/2<0

\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow-12+\sqrt{x}+3< 0\)

=>0<x<81 và x<>9