ta có: \(x^2+xy-2012x-2013y-2014=0.\)
\(\Leftrightarrow x\left(x+y\right)-2013\left(x+y\right)+x-2013=1\)
\(\Leftrightarrow\left(x+y\right)\left(x-2013\right)+x-2013=1\)
\(\Leftrightarrow\left(x-2013\right)\left(x+y+1\right)=1\)
mà x,y là các số nguyên nên
\(\orbr{\begin{cases}\hept{\begin{cases}x-2013=1\\x+y+1=1\end{cases}}\\\hept{\begin{cases}x-2013=-1\\x+y+1=-1\end{cases}}\end{cases}\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2014\\y=-2014\end{cases}}\\\hept{\begin{cases}x=2012\\y=-2012\end{cases}}\end{cases}}}\)
vậy (x;y)={ (2014;-2014) ;(2012;-2012)}
\(x^2+xy-2012x-2013y-2014=0\) \(0\)
\(\Leftrightarrow x\left(x+y\right)-2013x-2013y+x-2013-1=0\)
\(\Leftrightarrow x\left(x+y\right)-2013\left(x+y\right)+\left(x-2013\right)=1\)
\(\Leftrightarrow\left(x+y\right).\left(x-2013\right)+\left(x-2013\right)=1\)
\(\Leftrightarrow\left(x-2013\right).\left(x+y+1\right)=1\)
Mà x,y lại là số nguyên
Vậy \(\hept{\begin{cases}\left(x;y\right)=\left(2014;2014\right)\\\left(x;y\right)=\left(2012;2012\right)\end{cases}}\)
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