bạn ơi đề khó nhìn vậy  

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

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

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

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

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

<=>x²(x+y)+y²(x+y)=2001

<=>(x+y)(x²+y²)=2001

=>x+y, x²+y² E Ư(2001)={1;3;23;29;69;87;667;2001}

Rồi xét các trường hợp => x,y

\(pt\Leftrightarrow\left(x-1\right)\left(x-2y^2-y+2\right)=1\)

Ok ?!

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

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

\(\Leftrightarrow x-y=3\) 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 \(1-xy=0\)

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

ban oi tai sao den buoc 3 ban lai suy ra nhu vay duoc

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.........