a) de (x+1).(2y-)=12
thì x+1 va 2y-1 cung dau
\(a\Leftrightarrow\hept{\begin{cases}2x+1=\left\{-5,-1,1,5\right\}=>x=\left\{-3,-1,0,2\right\}\\y-3=\left\{-2,-10,10,2\right\}=>y=\left\{1,-7,13,5\right\}\end{cases}}\)
\(b\Leftrightarrow\hept{\begin{cases}x+1=\left\{-4,-12,12,4\right\}=>x=\left\{-5,-13,11,3\right\}\\2y-1=\left\{-3,-1,1,3\right\}=>y=\left\{-1,0,1,2\right\}\end{cases}}\)
\(c\Leftrightarrow\orbr{\begin{cases}y+1=0=>y=-1;x=\left\{R\right\}\\x-1=0=>x=1;y=\left\{R\right\}\end{cases}}\)
\(d\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)