Ta có : xy + x + y = -1
=> x(y + 1) + y + 1 = -1 + 1
=> (x + 1)(y + 1) = 0
=> \(\orbr{\begin{cases}x+1=0\\y+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\)(đpcm)
Vậy nếu xy + x + y = - 1 thì có ít nhất 1 số bằng - 1
xy + x + y = -1
<=> xy + x + y + 1 = 0
<=> x( y + 1 ) + 1( y + 1 ) = 0
<=> ( x + 1 )( y + 1 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\y+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\) ( đpcm )
\(xy+x+y=-1\)
\(< =>xy+x+y+1=0\)
\(< =>x\left(y+1\right)+\left(y+1\right)=0\)
\(< =>\left(x+1\right)\left(y+1\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\y+1=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\)ez