\(A=\left|x+2\right|+\left|x+5\right|+\left|x-7\right|+\left|x-8\right|\)
\(A=\left|x+2\right|+\left|x+5\right|+\left|7-x\right|+\left|8-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A\ge\left|x+2+x+5+7-x+8-x\right|=22\)
Dấu "=" xảy ra khi: \(\left(x+2\right)\left(x+5\right)\left(7-x\right)\left(8-x\right)\ge0\)
Nên \(\left\{{}\begin{matrix}x\ge-2\\x\ge-5\\x\le7\\x\le8\end{matrix}\right.\Leftrightarrow-2\le x\le7\)