Question

[s111111111]

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

Timf x nguyên để A nguyên

Answer 1

[Nguyễn Duy Khang]

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\\ A\in Z\Rightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\Rightarrow\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)\\ \Leftrightarrow\left(\sqrt{x}-1\right)\in\left\{2;1;-1;-2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;2;0;-1\right\}\\ \Leftrightarrow x\in\left\{9;4;0\right\}\)

Vậy \(x\in\left\{9;4;0\right\}\)