Question

Giải phương trình: \(x^2+xy-2008x-2009y-2010=0\)

Answer 1

Ta có:

\(x^2+xy-2008x-2009y-2010=0\)

\(\Leftrightarrow x^2+xy+x-2009x-2009y-2009=1\)

\(\Leftrightarrow x\left(x+y+1\right)-2009\left(x+y+1\right)=1\)

\(\Leftrightarrow\left(x+y+1\right)\left(x-2009\right)=1\)

Xét trường hợp:

\(\left(1\right)\left\{{}\begin{matrix}x-2009=1\\x+y+1=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2010\\y=-2010\end{matrix}\right.\)

\(\left(2\right)\left\{{}\begin{matrix}x-2009=-1\\x+y+1=-1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2008\\y=-2010\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2010;-2010\right);\left(2008;-2010\right)\)