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a)
\(xy+2x+2y=3\)
=> \(xy+2x+2y+4=7\)
=> \(x\left(y+2\right)+2\left(y+2\right)=7\)
=> \(\left(x+2\right)\left(y+2\right)=7\)
=> \(\left\{{}\begin{matrix}x+2=1\\y+2=7\end{matrix}\right.\)=> \(x=-1;y=5\)
=> \(\left\{{}\begin{matrix}x+2=7\\y+2=1\end{matrix}\right.\)=> \(x=5;y=-1\)
=> \(\left\{{}\begin{matrix}x+2=-1\\y+2=-7\end{matrix}\right.\)=> \(x=-3;y=-9\)
=> \(\left\{{}\begin{matrix}x+2=-7\\y+2=-1\end{matrix}\right.\)=> \(x=-9;y=-3\)
b)
\(y\left(x+1\right)=3x+5\)
=> \(xy+y=3x+5\)
=> \(xy+y-3x-5=0\)
=> \(x\left(y-3\right)+y-5=0\)
=> \(x\left(y-3\right)+y-3=2\)
=> \(\left(x+1\right)\left(y-3\right)=2\)
=> \(\left\{{}\begin{matrix}x+1=1\\y-3=2\end{matrix}\right.\)=> \(x=0;y=5\)
=> \(\left\{{}\begin{matrix}x+1=2\\y-3=1\end{matrix}\right.\)=> \(x=1;y=4\)
=> \(\left\{{}\begin{matrix}x+1=-1\\y-3=-2\end{matrix}\right.\)=> \(x=-2;y=1\)
=> \(\left\{{}\begin{matrix}x+1=-2\\y-3=-1\end{matrix}\right.\)=> \(x=-3;y=2\)
Ta có : xy - 3x + y =3
x(y - 3) + y - 3 = 0
(y - 3)(x+1) = 0
=> y - 3 = 0 hoặc x + 1 = 0
Còn lại bạn tự giải nhé
Ta có :
\(xy+2x+2y=3\)
\(\Leftrightarrow xy+2x+2y+7=3+4\)
\(\Leftrightarrow x\left(y+2\right)+2\left(y+2\right)=7\)
\(\Leftrightarrow\left(y+2\right)\left(x+2\right)=7\)
Vì \(x;y\in Z\Leftrightarrow y+2;x+2\in Z\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\x+2=7\end{matrix}\right.\Leftrightarrow y=-1;x=5\) \(\left(tm\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=7\\x+2=1\end{matrix}\right.\)\(\Leftrightarrow\begin{matrix}y=5;x=-1\\\end{matrix}\)\(\left(tm\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=-7\\x+2=-1\end{matrix}\right.\)\(\Leftrightarrow y=-9;x=-3\)\(\left(tm\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=-1\\x+2=-7\end{matrix}\right.\)\(\Leftrightarrow y=-3;x=-9\) \(\left(tm\right)\)
Vậy .....................