Question

tim x y biet x+y+xy+1=0

Answer 1

\(x+y+x.y+1=0\)

\(x.1+x.y+y+1\) \(=0\)

\(x.\left(1+y\right)+\left(y+1\right)\) \(=0\)

\(\left(1+y\right).\left(x+1\right)=0\)

\(\Rightarrow1+y=0\) \(\Rightarrow\) \(y=-1\)

\(\Rightarrow\) \(x+1=0\) \(\Rightarrow\) \(x=-1\)