Question

Cho biểu thức

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x-9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\) với \(x\ge0,x\ne9\)

a, Rút gọn C

b,Tìm x sao cho C < -1

Answer 1

a:

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

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

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

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

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

b: Để C<-1 thì C+1<0

=>-3 căn x+2 căn x+4<0

=>-căn x<-4

=>x>16