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\(A=\left(x+3y-5\right)^2-6xy+26\)
\(=x^2+9y^2+25+6xy-10x-30y-6xy+26\)
\(=x^2-10x+25+9y^2-30y+25+1\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\)
Vì :
\(\left(x-5\right)^2\ge0\forall x\)
\(\left(3y-5\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)
Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-5\right)^2=0\\\left(3y-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}\)
Vậy \(A_{min}=1\) tại \(\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}\)
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\(A=\left(x+3y-5\right)^2-6xy+26\)
\(=x^2+9y^2+25+6xy-30y-10x+26-6xy\)
\(=x^2+9y^2-10x+51-30y\)
\(=\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)
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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A = \(x^2+9y^2+25+6xy-30y-10x-6xy+26\)
= \(x^2-10x+25+9y^2-30y+25+1\)
= \(\left(x-5\right)^2+\left(3y-5\right)^2+1\)
Có : \(\left(x-5\right)^2\ge0\forall x;\left(3y-5\right)^2\ge0\forall y\)
\(\Rightarrow A\ge1\)
Vậy GTNN của A là 1 \(\Leftrightarrow\hept{\begin{cases}x-5=0\\3y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}}\)
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