\(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+4^2\right).\left[\left(3y\right)^2-2.3y.2+2^2\right]=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(3y-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{matrix}x+4=0\\3y-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-4\\y=\frac{2}{3}\end{matrix}\right.\)
Vậy ............
\(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right).\left(9y^2-6y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2.\left(3y-2\right)^2=0\)
\(\Leftrightarrow\left(x+4\right)\left(3y-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+4=0\\3y-2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-4\\y=\frac{2}{3}\end{matrix}\right.\)
Vậy \(x=-4\) và \(y=\frac{2}{3}\)
