a) \(xy+2x+3y=-6\)
\(\Rightarrow x\left(y+2\right)+3y+6=0\)
\(\Rightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(x=-3;y=-2\)
\(xy+2x+3y=-6\)
\(\Leftrightarrow xy+2x+3y+6=0\)
\(\Leftrightarrow y\left(x+3\right)+\text{2}\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Leftrightarrow\left\{\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
\(xy+2x+3y=-6\)
\(\Leftrightarrow x\left(y+2\right)+3y+6=-6+6\)
\(\Leftrightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Ta có xy+2x+3y=-6
\(\Leftrightarrow xy+2x+3y+6=0\)
\(\Leftrightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)
Với x+3=0 thì x=-3 và y tùy ý
Với y+2=0 thì y=-2 và x tùy ý
Vậy cặp (x;y) là (-3;tùy ý);(tùy ý;-2)
