Question

Tìm nghiệm nguyên:

y(x-1)=x²+2

Answer 1

\(x=1\) không phải là 1 nghiệm của pt

\(x\ne1\Rightarrow y=\frac{x^2+2}{x-1}=x+1+\frac{3}{x-1}\)

Để \(y\) nguyên \(\Rightarrow\frac{3}{x-1}\) nguyên \(\Rightarrow x-1=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(x-1=-3\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\)

\(x-1=-1\Rightarrow\left\{{}\begin{matrix}x=0\\y=-2\end{matrix}\right.\)

\(x-1=1\Rightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

\(x-1=3\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)