Question

Cho P=\(\left(\dfrac{x-6}{x+3\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{2x-6}{x+1}\)

a) Rút gọn P

b) Tìm x để P=1

Answer 1

a: \(P=\dfrac{x-6-\sqrt{x}-3+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\cdot\dfrac{x+1}{2x-6}\)

\(=\dfrac{x-9}{\sqrt{x}\left(\sqrt{x}+3\right)}\cdot\dfrac{x+1}{2x-6}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}}\cdot\dfrac{x+1}{2x-6}\)

b: Để P=1 thì \(\sqrt{x}\left(2x-6\right)=\left(x+1\right)\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow2x\sqrt{x}-6\sqrt{x}-x\sqrt{x}+3x-\sqrt{x}+3=0\)

\(\Leftrightarrow x\sqrt{x}+3x-7\sqrt{x}+3=0\)

=>x=1