Question

Cho P=\(x-\sqrt{x}+1\) với x>0; x\(\ne1\). Tìm x để \(Q=\dfrac{2\sqrt{x}}{P}\) nguyên

Answer 1

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

Q nguyên \(< =>\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

đặt \(\sqrt{x}+\dfrac{1}{\sqrt{x}}=t\left(t>0\right)=>t-1\inƯ\left(2\right)=>t\in\left\{2;0;3;-1\right\}\)

\(=>t\in\left\{2;0;3\right\}=>\sqrt{x}+\dfrac{1}{\sqrt{x}}\in\left\{2;0;3\right\}\)

\(=>x\in\left\{1;\right\}\)(loại ) vậy\(x\in\varnothing\)

(không chắc)