Lời giải:
$2x^2+y^2+2xy-6x-2y=8$

$\Leftrightarrow (x^2+y^2+2xy)+x^2-6x-2y=8$
$\Leftrightarrow (x+y)^2-2(x+y)+x^2-4x=8$

$\Leftrightarrow (x+y)^2-2(x+y)+1+(x^2-4x+4)=13$

$\Leftrightarrow (x+y-1)^2+(x-2)^2=13$
$\Rightarrow (x-2)^2=13-(x+y-1)^2\leq 13$
Mà $(x-2)^2$ là scp với mọi $x$ nguyên nên $(x-2)^2\in\left\{0; 1; 4; 9\right\}$

Nếu $(x-2)^2=0\Rightarrow (x+y-1)^2=13-(x-2)^2=13$ (không là scp - loại) 

Nếu $(x-2)^2=1\Rightarrow (x+y-1)^2=12$ (không là scp - loại)

Nếu $(x-2)^2=4\Rightarrow (x+y-1)^2=9$

$\Rightarrow x-2=\pm 2$ và $x+y-1=\pm 3$
TH1: $x-2=2; x+y-1=3\Rightarrow x=4; y=0$

TH2: $x-2=2; x+y-1=-3\Rightarrow x=4; y=-6$

TH3: $x-2=-2; x+y-1=3\Rightarrow x=0; y=4$

TH4: $x-2=-2; x+y-1=-3\Rightarrow x=0; y=-2$

Nếu $(x-2)^=9\Rightarrow (x+y-1)^2=4$ (bạn cũng làm tương tự trên)

Sửa đề: Tìm cặp \(x,y\in Z\) thỏa mãn \(x^2+3xy+2y^2+3x+6y-4=0\).

\(x^2+3xy+2y^2+3x+6y-4=0\)

\(\Leftrightarrow x^2+2xy+xy+2y^2+3x+6y=4\)

\(\Leftrightarrow\left(x^2+2xy\right)+\left(xy+2y^2\right)+\left(3x+6y\right)=4\)

\(\Leftrightarrow x\left(x+2y\right)+y\left(x+2y\right)+3\left(x+2y\right)=4\)

\(\Leftrightarrow\left(x+2y\right)\left(x+y+3\right)=4\)

Vì \(x,y\in Z\Rightarrow\left(x+2y\right)\left(x+y+3\right)\in Z\)

Trường hợp 1: \(\left\{{}\begin{matrix}x+2y=1\\x+y+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\) (thỏa mãn)

Trường hợp 2: \(\left\{{}\begin{matrix}x+2y=4\\x+y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=6\end{matrix}\right.\) (thỏa mãn)

Trường hợp 3: \(\left\{{}\begin{matrix}x+2y=2\\x+y+3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\) (thỏa mãn)

Trường hợp 4: \(\left\{{}\begin{matrix}x+2y=-2\\x+y+3=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=3\end{matrix}\right.\) (thỏa mãn)

Vậy: \(\left(x,y\right)=\left[\left(1;0\right),\left(-8;6\right),\left(-4;3\right),\left(-8;3\right)\right]\)

\(3x^2+3xy-17=7x-2y\)

\(\Leftrightarrow3x\left(x+y\right)+2x+2y-9x-17=0\)

\(\Leftrightarrow3x\left(x+y\right)+2\left(x+y\right)-9x-6-11=0\)

\(\Leftrightarrow\left(x+y\right)\left(3x+2\right)-3\left(3x+2\right)=11\)

\(\Leftrightarrow\left(3x+2\right)\left(x+y-3\right)=11\)

\(\Leftrightarrow\left(3x+2\right);\left(x+y-3\right)\in\left\{-1;1;-11;11\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(-\dfrac{1}{3};\dfrac{43}{3}\right);\left(-\dfrac{11}{3};\dfrac{17}{3}\right);\left(3;1\right)\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(3;1\right)\right\}\left(x;y\inℤ\right)\)

\(3xy+x+15y-44=0\)

\(3y\left(x+5\right)+\left(x+5\right)-49=0\)

\(\left(x+5\right)\left(3y+1\right)=49\)

Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)

Có \(\left(x+5\right)\left(3y+1\right)=49\)

\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)

b tự lập bảng nhé~

2x\(^2\)+y\(^2\)+3xy+3x+2y+2=0

\(\Leftrightarrow\)16x\(^2\)+8y\(^2\)+24xy+24x+16y+16=0

\(\Leftrightarrow\)(4x)\(^2\)+24x(y+1)+8y\(^2\)+16y+16=0

\(\Leftrightarrow\)(4x)\(^2\)+24x(y+1)+[3(y+1)]\(^2\)-[3(y+1)]\(^2\)+8y\(^2\)+16y+16=0

\(\Leftrightarrow\)(4x+3y+3)\(^2\)-9y\(^2\)-18y-9+8y\(^2\)16y+16=0

\(\Leftrightarrow\)(4x+3y+3)\(^2\)-y\(^2\)-2y-1+8=0

\(\Leftrightarrow\)(4x+3y+3)\(^2\)- (y+1)\(^2\)= -8

\(\Leftrightarrow\)(y+1+4x+3y+3) (y+1-4x-3y-3)=8

\(\Leftrightarrow\)4(x+y+4) (-4-2y-2)=8

\(\Leftrightarrow\)(x+y+4) (2x+y+11)= -1

\(\Leftrightarrow\){x+y+4= -1

      {2x+y+1=1

\(\Rightarrow\)x=2 và y= -4

{x+y+4= 1

{2x+y+1= -1

\(\Rightarrow\)x=-2 và y=2

vậy nghiệm (x,y)=(-2;4) (-2;2)

\(\Leftrightarrow y\left(3x+2\right)=7x+17-3x^2\)

Dễ thấy \(3x+2\ne0\)

\(\Leftrightarrow y=\frac{7x+17-3x^2}{3x+2}=-x+3+\frac{11}{3x+2}\)

Dể y nguyên thì \(3x+2\)phải là ước nguyên của 11

\(\Rightarrow3x+2=\left\{-11;-1;1;11\right\}\)