Question

Cho 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 đkxđ và rút gọn

b) Tìm x để C< -1

Answer 1

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

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

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

Answer 2

ĐKXĐ: x\(\ge0;x\ne3;x\ne-3\)

Answer 3

C<-1

\(\Leftrightarrow\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}< -1\)

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

\(\Leftrightarrow\dfrac{-\sqrt{x}+4}{2\left(\sqrt{x}+2\right)}< 0\)

Ta có: 2(\(\sqrt{x}+2)\ge0\)với mọi..........

nên -\(\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\)