Question

cho biểu thức

P=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)với x>0,x\(\ne\)9

a, rút gon P

b,tìm x sao cho P<-1

Answer 1

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

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

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

b: Để P<-1 thì P+1<0

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

=>0<x<16 và x<>9