Question

Tìm giá trị nhỏ nhất

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

Answer 1

Do \(\left|x-\dfrac{1}{2}\right|\ge0\forall x\) ; \(\left(y+2\right)^2\ge0\forall y\)

suy ra \(\left|x-\dfrac{1}{2}\right|+\left(y+2\right)^2\ge11\forall x;y\)

hay \(C\ge11\forall x;y\)

Dấu bằng xảy ra khi và chỉ khi:

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

Vậy....