\(C=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)
\(\ge0+0+2017=2017.\Rightarrow C_{min}=2017\Leftrightarrow\hept{\begin{cases}x-1=0\\x-3y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}\)
C= (x-3y)2+(x-1)2+2017 \(\ge2017\)
Min C = 2017
\(C=2x^2+9y^2-6xy-2x+2018\)
\(=x^2-6xy+9y^2+x^2-2x+1+2017\)
\(=\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\forall x;y\)
Dấu"=" xảy ra<=> \(\hept{\begin{cases}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x-3y=0\\x=1\end{cases}\Rightarrow y=\frac{1}{3}}\)
\(C=2x^2+9y^2-6xy-2x+2018\)
= \(\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017\)
= \(\left(x-3y\right)^2+\left(x-1\right)^2+2017\)
Nhận xét :
\(\hept{\begin{cases}\left(x-3y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\)
\(\Leftrightarrow C\ge2017\)
Dấu "= " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}\)
Vậy \(C_{min}=2017\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}\)
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