Question

Tìm GTNN

Answer 1

\(B=\left|x-\dfrac{1}{2}\right|+\left(y+2\right)^2+11\ge0+0+11=11\)

\(\Rightarrow minB=11\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-2\end{matrix}\right.\)

Answer 2

\(\left|x-\dfrac{1}{2}\right|+\left(y+2\right)^2+11\ge11\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{1}{2};-2\right)\)