\(A=\dfrac{x^2-1}{2x^2+1}\)
Để \(A\in Z\) thì: \(x^2-1⋮2x^2+1\)
\(\Rightarrow2x^2-2⋮2x^2+1\)
\(\Rightarrow2x^2+1-3⋮2x^2+1\)
\(\Rightarrow3⋮2x^2+1\)
\(\Rightarrow2x^2+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Dễ thấy: \(2x^2+1>0\) và \(2x^2+1\) lẻ
\(\Rightarrow\left[{}\begin{matrix}2x^2+1=1\\2x^2+1=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x^2=0\\2x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Đặt \(P=\dfrac{x^2-1}{2x^2+1}\)
\(P\in Z\Leftrightarrow x^2-1⋮2x^2+1\Leftrightarrow2x^2-2⋮2x^2+1\)
\(\Leftrightarrow2x^2+1-3⋮2x^2+1\Leftrightarrow3⋮2x^2+1\Leftrightarrow2x^2+1\inƯ\left(3\right)\)
\(\Rightarrow2x^2+1\in\left\{\pm1;\pm3\right\}\)
Do \(2x^2+1>0\Rightarrow2x^2+1\in\left\{1;3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2x^2+1=1\\2x^2+1=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x^2=0\\2x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^{ }=0\\x^{ }=\pm1\end{matrix}\right.\)
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