\(x^2y^2-16xy+99=9x^2+36y^2+13x+26y\)

\(\Leftrightarrow\left(xy+10\right)^2=9\left(x+2y\right)^2+13\left(x+2y\right)+1\)

Khi đó ta dễ thấy:

\(\left(3x+6y\right)^2< \left(xy+10\right)^2< \left(3x+6y+2\right)^2\)

\(\Rightarrow\left(xy+10\right)^2=\left(3x+6y+1\right)^2\)

Đến đây thì quá dễ rồi nhá, bạn tự làm nốt

\(\Leftrightarrow\left(x^2y^4-16xy^3+64y^2\right)+\left(4y^2-4xy+x^2\right)=0\)

\(\Leftrightarrow\left(xy^2-8y\right)^2+\left(2y-x\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy^2-8y=0\\2y-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy^2-8y=0\\x=2y\end{matrix}\right.\)

\(\Rightarrow2y.y^2-8y=0\)

\(\Leftrightarrow2y\left(y^2-4\right)=0\Rightarrow\left[{}\begin{matrix}y=0\Rightarrow x=0\\y=2\Rightarrow x=4\\y=-2\Rightarrow x=-4\end{matrix}\right.\)

\(\Leftrightarrow x^2y^2+22xy+141=4\left(x^2+6xy+9y^2\right)+7\left(x+3y\right)\)

\(\Leftrightarrow\left(xy+11\right)^2+20=4\left(x+3y\right)^2+7\left(x+3y\right)\)

\(\Leftrightarrow16\left(xy+11\right)^2+320=64\left(x+3y\right)^2+112\left(x+3y\right)\)

\(\Leftrightarrow\left(4xy+44\right)^2+369=\left(8x+24y+7\right)^2\)

\(\Leftrightarrow\left(8x+24y-4xy-37\right)\left(8x+24y+4xy+51\right)=369\)

Pt ước số

\(hpt\Leftrightarrow\hept{\begin{cases}xy=y^2\\x^2+y^2=-50\end{cases}}\)

Dễ thấy: \(VT=x^2+y^2\ge0>-50=VP\)

sai đề

7y^2  giúp cái tui cx mắc ở bài này