Question

rút gọn biểu thức của p=\(\left(\dfrac{1}{x+\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}\right)\):\(\dfrac{2}{\sqrt{x}+1}\)với x>0

Answer 1

Answer 2

 Ta có \(P=\left(\dfrac{1}{x+\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2}{\sqrt{x}+1}\)

\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2}{\sqrt{x}+1}\)

\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{2}\)

\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\) 

Answer 3

Ta có: \(P=\left(\dfrac{1}{x+\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2}{\sqrt{x}+1}\)

\(=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\)

\(=\dfrac{x+1}{2\sqrt{x}}\)