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

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

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

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

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

bạn ơi đề khó nhìn vậy  

<=>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

Ta có:

x(x2+x+1)=4y(y+1)x(x2+x+1)=4y(y+1)

⟺x3+x2+x+1=4y2+4y+1⟺x3+x2+x+1=4y2+4y+1

⟺(x2+1)(x+1)=(2y+1)2⟺(x2+1)(x+1)=(2y+1)2 (*)

Đặt (x2+1;x+1)=d(x2+1;x+1)=d

⟹(x+1)(x−1)−(x2+1)⋮d⟹(x+1)(x−1)−(x2+1)⋮d

⟹2⋮d⟹2⋮d

Dễ thầy VPVP của phương trình (∗)(∗) là số lẻ nên chỉ xảy ra trường hợp d=±1d=±1

⟹x2+1=a2⟹x2+1=a2 và x+1=b2x+1=b2

Từ đây dễ dàng suy ra x=0x=0

⟹y=0;y=−1⟹y=0;y=−1

Thử lại ta thấy (x;y)=(0;0);(0;−1)(x;y)=(0;0);(0;−1)

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

Ok ?!