Ta có:
\(xy+2x+y+11=0\\ \Rightarrow xy+2x+y+11+2=0+2\\ \Rightarrow x\left(y+2\right)+y+2=2-11\\ \Rightarrow\left(y+2\right)\left(x+1\right)=-9\)
\(\Rightarrow\left(y+2\right)\left(x+1\right)=1.-9\) hoặc \(\left(y+2\right)\left(x+1\right)=-1.9\)
TH1: \(y+2=1;x+1=-9\)
\(y+2=1\Rightarrow y=1-2\\\Rightarrow y=-1\)
\(x+1=-9\Rightarrow x=-9-1\\ \Rightarrow x=-10\)
TH2: \(y+2=-9;x+1=1\)
\(y+2=-9\Rightarrow y=-9-2\\ \Rightarrow y=-11\)
\(x+1=1\Rightarrow x=1-1\\ \Rightarrow x=0\)
TH3: \(y+2=-1;x+1=9\)
\(y+2=-1\Rightarrow y=-1-2\\ \Rightarrow y=-3\)
\(x+1=9\Rightarrow x=9-1\\ \Rightarrow x=8\)
TH4: \(y+2=9;x+1=-1\)
\(y+2=9\Rightarrow y=9-2\\ \Rightarrow y=7\)
\(x+1=-1\Rightarrow x=-1-1\\ \Rightarrow x=-2\)
Vậy: \(\left(x;y\right)\in\left\{\left(-10;-1\right);\left(0;-11\right);\left(8;-3\right);\left(-2;7\right)\right\}\)
