Question

0=(x+y)(x+1)(y+1) biết xy=2
                              x+y+1=0

Answer 1

\(Q=...=-1\left[x+y+xy+1\right]=-1\left[-1+2+1\right]=-1.2=-2\)

Answer 2

\(Q=\left(x+y\right)\left(x+1\right)\left(y+1\right)\) biết \(xy=2\) ; \(x+y+1=0\) 

Giải:

Ta có:

\(x+y+1=0\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=-1\\x+1=-y\\y+1=-x\end{matrix}\right.\) 

Từ đó, ta có:

\(Q=\left(x+y\right)\left(x+1\right)\left(y+1\right)=\left(-1\right)\left(-y\right)\left(-x\right)\) 

mà \(xy=2\) 

\(\Rightarrow Q=\left(-1\right).2=-2\)