Question

Tìm x để :

A=\(\frac{2\sqrt{x}}{x-\sqrt{x}+1}\) ∈ Z

Answer 1

ĐK:\(x\ge0\)

\(A\in Z\Rightarrow2\sqrt{x}⋮x-\sqrt{x}+1\left(1\right)\)

\(\Leftrightarrow2x⋮x-\sqrt{x}+1\)

\(\Leftrightarrow2x-2\sqrt{x}+2+2\sqrt{x}-2⋮x-\sqrt{x}+1\)

\(\Rightarrow2\sqrt{x}-2⋮x-\sqrt{x}+1\left(2\right)\)

Từ (1)(2), có:\(2\sqrt{x}-\left(2\sqrt{x}-2\right)=2⋮x-\sqrt{x}+1\)\(\Rightarrow x-\sqrt{x}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x\in\left\{0\right\}\)