\(A=x^2+5y^2+4xy+2x+12\)
\(A=x^2+4xy+4y^2+2\left(x+2y\right)+1+y^2-4y+4+7\)
\(A=\left(x+2y\right)^2+2\left(x+2y\right)+1+\left(y-2\right)^2+7\)
\(A=\left(x+2y+1\right)^2+\left(y-2\right)^2+7\)
Vì \(\left(x+2y+1\right)^2\ge0\forall x;y\)và \(\left(y-2\right)^2\ge0\forall y\)
\(\Rightarrow A\ge7\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=2\end{cases}}}\)
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