Question

Cho    \(^{\dfrac{x\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{x-1}{\sqrt{x}-1}}\)  ( Đk x ≥ 0,x≠1)

giải

 

Answer 1

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