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

\(\left(2x-3\right)^2+4x^2-9=\left(2x-3\right)\left(3x+5\right)\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(2x-3\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+5\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left[\left(2x-3\right)+\left(2x+3\right)-\left(3x+5\right)\right]=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3+2x+3-3x-5\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};5\right\}\)