Question

Tìm GTNN:

d) D = \(\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{5}{3}\right|+\left|x-3\right|\).

Answer 1

Giải:

Ta biết rằng \(\left\{{}\begin{matrix}\left|A\right|\ge A\Leftrightarrow A\ge0\\\left|A\right|\ge0\Leftrightarrow A=0\end{matrix}\right.\)

Ta có: \(D=\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{5}{3}\right|+\left|x-3\right|\)

\(=\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{5}{3}\right|+\left|3-x\right|\) \(\ge x-\dfrac{1}{2}+0+3-x=\dfrac{5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}\ge0\\x-\dfrac{5}{3}=0\\3-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{5}{3}\\x\le3\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(MIN_D=\dfrac{5}{2}\Leftrightarrow x=\dfrac{5}{3}\)