1, \(4x^2+25y^2-12x-20y+13=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+\left(25y^2-20y+4\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2+\left(5y-2\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(2x-3\right)^2\ge0\\\left(5y-2\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2+\left(5y-2\right)^2\ge0\)
Mà \(\left(2x-3\right)^2+\left(5y-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=0\\\left(5y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{2}{5}\end{matrix}\right.\)
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b, \(13x^2+y^2+4xy-34x-2y+26=0\)
\(\Leftrightarrow\left(4x^2+y^2+1+4xy-4x-2y\right)+9x^2-30x+25=0\)
\(\Leftrightarrow\left(2x+y-1\right)^2+\left(3x-5\right)^2=0\)
Vì mỗi nhóm \(\ge0\) mà tổng 2 nhóm trên = 0
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y-1\right)^2=0\\\left(3y-5\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-7}{3}\\x=\dfrac{5}{3}\end{matrix}\right.\)
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