Để \(\dfrac{a^2+a+3}{a+1}\in Z\) thì :
\(a^2+a+3⋮a+1\)
Mà \(a+1⋮a+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+a+3⋮a+1\\a^2+a⋮a+1\end{matrix}\right.\)
\(\Leftrightarrow3⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}a+1=1\\a+1=-1\\a+1=3\\a+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-2\\a=2\\a=-4\end{matrix}\right.\)
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\(x-2xy+y=0\)
\(\Leftrightarrow2\left(x-2xy+y\right)=0\)
\(\Leftrightarrow2x-4xy+2y=0\)
\(\Leftrightarrow2x-4xy+2y-1=-1\)
\(\Leftrightarrow\left(2x-4xy\right)-\left(1-2y\right)=-1\)
\(\Leftrightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1=1\\1-2y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1=-1\\1-2y=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\end{matrix}\right.\)
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