Bài làm:
Ta có: \(4x^2+2y^2+4xy-4x-8y+15\)
\(=\left(4x^2+4xy+y^2\right)-2\left(2x+y\right)+1+y^2-6y+9+5\)
\(=\left(2x+y\right)^2-2\left(2x+y\right)+1+\left(y-3\right)^2+5\)
\(=\left(2x+y-1\right)^2+\left(y-3\right)^2+5\ge5\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x+y-1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy \(Min=5\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
4x2 + 2y2 + 4xy - 4x - 8y + 15
= [ ( 4x2 + 4xy + y2 ) - 2( 2x + y ) + 1 ] + ( y2 - 6y + 9 ) + 5
= ( 2x + y - 1 )2 + ( y - 3 )2 + 5
\(\hept{\begin{cases}\left(2x+y-1\right)^2\ge0\forall x,y\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(2x+y-1\right)^2+\left(y-3\right)^2+5\ge5\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x+y-1=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy GTNN của biểu thức = 5 <=> x = -1 ; y = 3
Ta có:\(4x^2+2y^2+4xy-4x-8y+15\)
\(=2\left(x^2+2xy+y^2\right)-8\left(x+y\right)+8+2x^2+4x+2+5\)
\(=2\left(x+y\right)^2-2.4\left(x+y\right)+2.4+2\left(x^2+2x+1\right)+5\)
\(=2\left(x+y-2\right)^2+2\left(x+1\right)^2+5\ge5\forall x,y\)
Dấu"=" xảy ra khi \(\orbr{\begin{cases}2\left(x+y-2\right)^2=0\\2\left(x+1\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\y=3\end{cases}}}\)
Vậy \(Min=5\)khi \(x=-1;y=3\)
