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\(x^2+y^2=3-xy\)

\(\Leftrightarrow\left(x-y\right)^2+2xy=3-xy\)

\(\Leftrightarrow\left(x-y\right)^2=3-3xy\)

\(\Leftrightarrow\left(x-y\right)^2=3\left(1-xy\right)\)

mà \(\left(x-y\right)^2\ge0,\forall x;y\inℤ\)

PT\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\1-xy=3\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-y=0\\1-xy=0\end{matrix}\right.\)

\(TH1:\left\{{}\begin{matrix}x-y=3\\1-xy=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\xy=-2\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;-2\right);\left(2;-1\right);\left(-1;2\right);\left(-2;1\right)\right\}\)

\(TH2:\left\{{}\begin{matrix}x-y=0\\1-xy=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\xy=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

Vậy \(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;-2\right);\left(2;-1\right);\left(-1;2\right);\left(-2;1\right);\left(1;1\right);\left(-1;-1\right)\right\}\)

\(x^2+y^2=3-xy\)

\(\Leftrightarrow\left(x-y\right)^2=3.\left(1-xy\right)\)

\(\Leftrightarrow x-y=3\) và \(1-xy=3\)

\(\Leftrightarrow\left(x;y\right)=\left(1;-2\right),\left(2;-1\right),\left(-1;2\right),\left(-2;1\right)\)

hoặc \(x-y=0\) và \(1-xy=0\)

\(\Leftrightarrow\left(x;y\right)=\left(1;1\right),\left(-1;-1\right)\)

Dễ

Lời giải:PT $\Leftrightarrow x^2+x(y-2014)-(2015y+2016)=0$

Coi đây là PT bậc 2 ẩn $x$. Để pt có nghiệm nguyên thì:

$\Delta=(y-2014)^2+4(2015y+2016)=t^2$ với $t\in\mathbb{N}$

$\Leftrightarrow y^2+4032y+4064260=t^2$

$\Leftrightarrow (y+2016)^2+4=t^2$$\Leftrightarrow 4=(t-y-2016)(t+y+2016)$

Đến đây thì đơn giản rồi thì đây là dạng phương trình tích.

2(x+y)+16-xy=0

<=> 2x+2y+16-xy=0

<=> y(2-x)-2(2-x)+20=0

<=> (2-x)(y-2)=-20

Vì x,y thuộc Z

=> 2-x;y-2 thuộc Z

=> 2-x;y-2 \(\inƯ\left(-20\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)

Xét bảng

Vậy.........

\(x^2+xy-2012x-2013y-2014=0\)

\(\Leftrightarrow x\left(x+y\right)-2013x-2013y+x-2013-1=0\)

\(\Leftrightarrow x\left(x+y\right)-2013\left(x+y\right)+\left(x-2013\right)-1=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2013\right)+\left(x-2013\right)-1=0\)

\(\Leftrightarrow\left(x-2013\right)\left(x+y+1\right)=1\)

\(\Leftrightarrow\left(x-2013\right);\left(x+y+1\right)\in\left\{-1;1\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(2012;-2014\right);\left(2014;-2014\right)\right\}\left(x;y\inℤ\right)\)