Ta có: A = \(\left(x+3y-5\right)^2-6xy+26\)
=> A = \(x^2+9y^2+25+6xy-10x-30y+6xy+26\)
=> A = \(\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)
=> A = \(\left(x-5\right)^2+\left(3y-5\right)^2+1\)
Vì \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left(3y-5\right)^2\ge0\end{matrix}\right.\) => A \(\ge\) 1
=> Dấu bằng xảy ra <=> \(\left\{{}\begin{matrix}x-5=0\\3y-5=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\y=\dfrac{5}{3}\end{matrix}\right.\)
=> GTNN của A =1 khi x = 5; y = \(\dfrac{5}{3}\)
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