Question

b1:

\(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)               với x>0

a,  rút gọn P

b,  tìm x để P=\(\dfrac{13}{3}\)

Answer 1

Answer 2

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

b) Để \(P=\dfrac{13}{3}\) thì \(3x+3\sqrt{x}+3-13\sqrt{x}=0\)

\(\Leftrightarrow3x-9\sqrt{x}-\sqrt{x}+3=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\)