a/ \(A=4x^2+y^2-4x-2y+3\)
\(=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
\(\Leftrightarrow A\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
Vậy...
b/ \(B=x^2+2y^2+2xy-2y\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(=\left(x+y\right)^2+\left(y-1\right)^2-1\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2-1\ge0\)
\(\Leftrightarrow B\ge-1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy..
a.\(A=4x^2+y^2-4x-2y+3\)
\(A=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(A=\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Vì \(\left(2x-1\right)^2\ge0\) và \(\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
\(\Rightarrow Min_A=1\) khi \(\left\{{}\begin{matrix}2x-1=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
b.\(B=x^2+2y^2+2xy-2y\)
\(B=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(B=\left(x+y\right)^2+\left(y-1\right)^2-1\)
Vì \(\left(x+y\right)^2\ge0\) và \(\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\)
\(\Rightarrow Min_B=-1\) khi \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy \(Min_B=-1\) khi \(x=-1;y=1\)
