\(M=2x^2+5y^2-2xy+2y+2x\)
\(2M=4x^2+10y^2-4xy+4y+4x\)
\(2M=\left(4x^2-4xy+y^2\right)+9y^2+4x+4y\)
\(2M=\left[\left(2x-y\right)^2+2\left(2x-y\right)+1\right]+\left(9y^2+6y+1\right)-2\)
\(2M=\left(2x-y+1\right)^2+\left(3y+1\right)^2-2\)
Do : \(\left(2x-y+1\right)^2\ge0\forall x;y\)
\(\left(3y+1\right)^2\ge0\forall y\)
\(\Rightarrow2M\ge-2\)
\(\Leftrightarrow M\ge-1\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}2x-y+1=0\\3y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{3}\\y=\frac{-1}{3}\end{cases}}\)
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