Question

Cho biểu thức :

A = \(\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}\) Với x > 0 ; x khác 1

a) Rút gọn A

b) Tìm x để A = 2017

Answer 1

a) \(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}\)

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

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

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

b) \(A=2017\)

\(\Rightarrow\dfrac{1+x}{1-x}=2017\)

\(\Rightarrow1+x=2017\left(1-x\right)\)

\(\Rightarrow1+x=2017-2017x\)

\(\Rightarrow2018x=2016\)

\(\Rightarrow x=\dfrac{1008}{1009}\)