\(A=\left(x+3y-5\right)^2-6xy+26\)
\(=x^2+9y^2+25+6xy-30y-10x+26-6xy\)
\(=x^2+9y^2-10x+51-30y\)
\(=\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\3y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{5}{3}\end{matrix}\right.\)
Vậy \(Min_A=1\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{5}{3}\end{matrix}\right.\)
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