Question

Cho

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

Tìm \(x\in R\) để \(A\) nguyên

Answer 1

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)

Để A \(\in Z\Rightarrow2⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(2\right)\)

\(\Rightarrow\sqrt{x}-3\in\left\{\pm1;\pm2\right\}\)

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Answer 2

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x-3}}=\dfrac{\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{2}{\sqrt{x-3}}=1+\dfrac{2}{\sqrt{x-3}}\)

\(\Rightarrow2⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(2\right)\)

\(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3+1=4\\\sqrt{x}=3+-1=2\\\sqrt{x}=3+2=5\\\sqrt{x}=3+-2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=4\\x=25\\x=1\end{matrix}\right.\)