\(\Leftrightarrow\left(x^2-2xy+y^2\right)+8\left(x-y\right)+16=3-2y^2\)

\(\Leftrightarrow\left(x-y\right)^2+8\left(x-y\right)+16=3-2y^2\)

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

Do \(\left(x-y+4\right)^2\ge0;\forall x,y\)

\(\Rightarrow3-2y^2\ge0\Rightarrow y^2\le\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}y^2=0\\y^2=1\end{matrix}\right.\)

\(\Rightarrow y=\left\{-1;0;1\right\}\)

- Với \(y=-1\) thay vào (1):

\(\left(x+5\right)^2=1\Rightarrow\left[{}\begin{matrix}x+5=1\\x+5=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\)

- Với \(y=1\) thay vào (1):

\(\Rightarrow\left(x+3\right)^2=1\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

- Với \(y=0\)

\(\Rightarrow\left(x+4\right)^2=3\) (ko có nghiệm nguyên do 3 ko phải SCP)

\(2y^2+2xy+x+3y-13=0\)

\(\Leftrightarrow2y\left(y+x\right)+x+y+2y=13\)

\(\Leftrightarrow\left(x+y\right)\left(2y+1\right)+2y+1=14\)

\(\Leftrightarrow\left(2y+1\right)\left(x+y+1\right)=14\)

Rồi bạn làm từng cặp ra nhé! 

VINSCHOOL

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

\(\Rightarrow2x-2y+4xy=6\)

\(\Rightarrow2x-2y+4xy-1=5\)

\(\Rightarrow\left(2x+4xy\right)-\left(2y+1\right)=5\)

\(\Rightarrow2x\left(2y+1\right)-1\left(2y+1\right)=5\)

\(\Rightarrow\left(2x-1\right)\left(2y+1\right)=5\)

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

\(\Leftrightarrow2\left(x-y+2xy\right)=2\times3\)

\(\Leftrightarrow2x-2y+4xy=6\)

\(\Leftrightarrow2x-2y+4xy-1=5\)

\(\Leftrightarrow\left(2x-4xy\right)-\left(2y+1\right)=5\)

\(\Leftrightarrow2x\left(2y+1\right)-\left(2y+1\right)=5\)

\(\Leftrightarrow\left(2x-1\right)\left(2y+1\right)=5\)

Bạn tự lập bảng để tìm nghiệm nhé

Gợi ý:
\(2xy+14x+y=33\)
\(\Rightarrow2x\left(y+7\right)+y+7=33+7\)
\(\Rightarrow\left(2x+1\right)\left(y+7\right)=40\)
\(\Rightarrow\left(2x+1;y+7\right)\inƯ\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Đến đây thì bạn làm tiếp nhé!

sao mình ko có cấp bậc?

=>2x(y+7)+y+7=40

=>(y+7)(2x+1)=40

mà x,y nguyên

nên \(\left(2x+1;y+7\right)\in\left\{\left(1;40\right);\left(5;8\right);\left(-1;-40\right);\left(-5;-8\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;33\right);\left(2;1\right);\left(-1;-47\right);\left(-3;-15\right)\right\}\)