Question

Tìm giá trị nhỏ nhất

A=|x+1|+|x-3|+5

B=|x-1|+|x-2|+|x-3|+10

C=|x-1|+|x-2|+|x-3|+|x-4|

D=\(\dfrac{6\left|y+5\right|+14}{2\left|y+5\right|+14}\)

Answer 1

Ta có: \(\left\{{}\begin{matrix}\left|x+1\right|\ge x+1\\\left|x-3\right|=\left|3-x\right|\ge3-x\end{matrix}\right.\)

\(\Rightarrow\left|x+1\right|+\left|3-x\right|+5\ge\left(x+1\right)+\left(3-x\right)+5\)

\(\Rightarrow\left|x+1\right|+\left|x-3\right|+5\ge9\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le3\end{matrix}\right.\)

\(\Leftrightarrow-1\le x\le3\)

Vậy Min_A = 9 \(\Leftrightarrow-1\le x\le3\)