Áp dụng BĐT \(|a|+\left|b\right|\ge\left|a+b\right|\):
\(A=\left|x-2\right|+\left|3-2x\right|+\left|4x-1\right|+\left|10-5x\right|\)
\(\ge\left|1-x\right|+\left|x-9\right|\ge\left|-8\right|=8\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}\left(x-2\right)\left(3-2x\right)\ge0\\\left(4x-1\right)\left(10-5x\right)\ge0\\\left(1-x\right)\left(x-9\right)\ge0\end{matrix}\right.\)\(\Leftrightarrow\dfrac{3}{2}\le x\le2\)
\(A=\left|x-2\right|+\left|2x-3\right|+\left|4x-1\right|+\left|5x-10\right|=\left|x-2\right|+\left|3-2x\right|+\left|1-4x\right|+\left|5x-10\right|\)\(A\ge x-2+3-2x+1-4x+5x-10=-8\)
vậy A\(\ge\)-8
