Question

giải pt nghiệm nguyên 

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

Answer 1

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

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

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

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

\(\Rightarrow\)\(\left(x-2009\right)=1\)và  \(\left(x+y+1\right)=1\)\(\Rightarrow\)\(x=2010;y=-2010\)

và     \(\left(x-2009\right)=-1\) và \(\left(x+y+1\right)=-1\)\(\Rightarrow\)\(x=2008;y=-2010\).