\(D=\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\)
\(D=\left|x-2\right|+\left|4-x\right|+\left|x-3\right|\)
Ta có:
\(\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|=2\)
Mà \(\left|x-3\right|\ge0\)
\(\Rightarrow D\ge2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-2\right)\left(4-x\right)\ge0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\le x\le4\\x=3\end{matrix}\right.\)
\(\Rightarrow x=3\)
Vậy \(Min_D=2\) khi x = 3.
Cho em thử sức nhé.Ko copy bạn dưới
\(LINH=\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\)
\(LINH=\left|x-2\right|+\left|x-4\right|+\left|x-3\right|\)
\(LINH=\left|x-2\right|+\left|4-x\right|+\left|x-3\right|\)
\(LINH\ge \left|x-2+4-x\right|+\left|x-3\right|\)
\(LINH\ge2+\left|x-3\right|\)
\(LINH\ge2\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-3=0\\x-4\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x=3\\x\le4\end{matrix}\right.\)
Vậy \(x=3\)
