Ta có: \(C=\left|x-\dfrac{2}{5}\right|+\left|x-\dfrac{3}{5}\right|=\left|x-\dfrac{2}{5}\right|+\left|\dfrac{3}{5}-x\right|\ge\left|x-\dfrac{2}{5}+\dfrac{3}{5}-x\right|=\left|\dfrac{1}{5}\right|=\dfrac{1}{5}\)hay C \(\ge\dfrac{1}{5}\Rightarrow C=\dfrac{1}{5}khi\left\{{}\begin{matrix}x-\dfrac{2}{5}\ge0\\\dfrac{3}{5}-x\ge0\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{2}{5}\\x\le\dfrac{3}{5}\end{matrix}\right.\)Vậy minC=\(\dfrac{1}{5}khi\dfrac{2}{5}\le x\le\dfrac{3}{5}\).....Chúc các bạn học tốt
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(C=\left|x-\dfrac{2}{5}\right|+\left|x-\dfrac{3}{5}\right|\)
\(=\left|x-\dfrac{2}{5}\right|+\left|-\left(x-\dfrac{3}{5}\right)\right|\)
\(=\left|x-\dfrac{2}{5}\right|+\left|-x+\dfrac{3}{5}\right|\)
\(\ge\left|x-\dfrac{2}{5}+\left(-x\right)+\dfrac{3}{5}\right|=\dfrac{1}{5}\)
Đẳng thức xảy ra khi \(\dfrac{2}{5}\le x\le\dfrac{3}{5}\)
