Ta có:

\(2010x^2+2011y^2-4020x+4022y+4021=0\)

\(\Leftrightarrow(2010x^2-4020x+2010)+(2011y^2+4022y+2011)=0\)

\(\Leftrightarrow2010(x^2-2x+1)+2011(y^2+2y+1)=0\)

\(\Leftrightarrow2010(x-1)^2+2011(y+1)^2=0\)

Ta thấy:

\(\left\{{}\begin{matrix}2010\left(x-1\right)^2\ge0\forall x\\2011\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow2010(x-1)^2+2011(y+1)^2\ge0\forall x,y\)

Mà \(\Leftrightarrow2010(x-1)^2+2011(y+1)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}2010\left(x-1\right)^2=0\\2011\left(y+1\right)^2=0\end{matrix}\right.\)

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

Vậy \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(2010x^2+2011y^2-4020x+4022y+4021=0\)

\(\Leftrightarrow\left(2010x^2-4020x+2010\right)+\left(2011y^2+4022y+2011\right)=0\)

\(\Leftrightarrow2010\left(x^2-2x+1\right)+2011\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\)

Vì \(2010\left(x-1\right)^2\ge0\forall x;2011\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow2010\left(x-1\right)^2+2011\left(y+1\right)^2\ge0\)

Để \(2010\left(x-1\right)^2+2011\left(y+1\right)^2=0\Leftrightarrow\hept{\begin{cases}2010\left(x-1\right)^2=0\\2011\left(y+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)

Vậy ..........

Lời giải:

PT $\Leftrightarrow x^3+x+1-y(x^2-3)=0$

$\Leftrightarrow y=\frac{x^3+x+1}{x^2-3}$ (hiển nhiên $x^2-3\neq 0$ với mọi $x$ nguyên) 

Để $y$ nguyên thì $\frac{x^3+x+1}{x^2-3}$ nguyên 

$\Leftrightarrow x^3+x+1\vdots x^2-3$
$\Rightarrow x(x^2-3)+4x+1\vdots x^2-3$
$\Rightarrow 4x+1\vdots x^2-3$

Hiển nhiên $4x+1\neq 0$ nên $|4x+1|\geq x^2-3$
Nếu $x\geq \frac{-1}{4}$ thì $4x+1\geq x^2-3$
$\Leftrightarrow x^2-4x-4\leq 0$

$\Leftrightarrow (x-2)^2\leq 8<9$

$\Rightarrow -3< x-2< 3$

$\Rightarrow -1< x< 5$

$\Rightarrow x\in \left\{0; 1; 2; 3; 4\right\}$.

Nếu $x< \frac{-1}{4}$ thì $-4x-1\geq x^2-3$

$\Leftrightarrow x^2+4x-2\leq 0$

$\Leftrightarrow (x+2)^2-6\leq 0$

$\Leftrightarrow (x+2)^2\leq 6< 9$

$\Rightarrow -3< x+2< 3$
$\Rightarrow -5< x< 1$

$\Rightarrow x\in\left\{-4; -3; -2; -1\right\}$

Đến đây bạn thay vào tìm $y$ thôi

\(\Leftrightarrow2x^2+x+2=y\left(2x-1\right)\)

\(\Leftrightarrow y=\dfrac{2x^2+x+2}{2x-1}=x+1+\dfrac{3}{2x-1}\)

\(y\in Z\Rightarrow\dfrac{3}{2x-1}\in Z\)

Mà x nguyên dương \(\Rightarrow2x-1>0\)

\(\Rightarrow2x-1=Ư\left(3\right)\Rightarrow x=\left\{1;2\right\}\) 

\(\Rightarrow\left(x;y\right)=\left(1;5\right);\left(2;4\right)\)