\(4x^2-4x+1+9y^2-6y+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
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Ta có:4x2-4x+9y2-6y+2=0
<=>(4x2-4x+1)+(9y2-6y+1)=0
<=> (2x-1)2+(3y-1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
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