Ta có: x(y+2)+y=1
\(\Leftrightarrow x\left(y+2\right)+y+2=1+2\)
\(\Leftrightarrow x\left(y+2\right)+\left(y+2\right)=3\)
\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=1\)=1.1=(-1).(-1)
\(\Rightarrow\left[{}\begin{matrix}x+1=1;y+2=1\\x+1=\left(-1\right);y+2=\left(-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-1=0;y=1-2=\left(-1\right)\\x=\left(-1\right)-1=\left(-2\right);y=\left(-1\right)-2=\left(-3\right)\end{matrix}\right.\)
Vậy: \(\left[{}\begin{matrix}x=0;y=\left(-1\right)\\z=\left(-2\right);y=\left(-3\right)\end{matrix}\right.\)
\(x\left(y+2\right)+y=1\\ \Leftrightarrow x\left(y+2\right)+1\left(y+2\right)=1+2\\ \Leftrightarrow\left(y+2\right)\left(x+1\right)=3\\ \cdot\left\{{}\begin{matrix}y+2=1\\x+1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\\ \cdot\left\{{}\begin{matrix}y+2=3\\x+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\\ \cdot\left\{{}\begin{matrix}y+2=-1\\x+1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=-4\end{matrix}\right.\\ \cdot\left\{{}\begin{matrix}y+2=-3\\x+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-2\end{matrix}\right.\)
Vậy ...
