Question

 Tìm x nguyên để \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) nguyên

Answer 1

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

A nguyên khi và chỉ khi:

\(\sqrt{x}-3\inƯ_4=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow\sqrt{x}\inƯ_4=\left\{1;2;4;5;7\right\}\)

\(\Leftrightarrow x\inƯ_4=\left\{1;4;16;25;49\right\}\)

Answer 2

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

hay \(x\in\left\{16;4;25;1;49\right\}\)