a ) \(yx^2+yx+y=1\)
\(\Leftrightarrow y\left(x^2+x+1\right)=1\)
Do x ; y nguyên \(\Rightarrow y;x^2+x+1\in Z\)
Mà \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow\left\{{}\begin{matrix}y=1\\x^2+x+1=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=1\\x\left(x+1\right)\end{matrix}\right.=0\Rightarrow\left\{{}\begin{matrix}y=1\\\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)
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b ) \(x^2+x-y^2=0\)
\(\Leftrightarrow x\left(x+1\right)=y^2\)
Do \(y^2\) là scp ; \(x\left(x+1\right)\)là tích 2 số nguyên liên tiếp
\(\Rightarrow\left\{{}\begin{matrix}y=0\\\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)
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