\(A=\left|x-1,5\right|+\left|x-2,5\right|\)
\(=\left|x-1,5\right|+\left|2,5-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1\)
Dấu "=" xảy ra khi \(1,5\le x\le2,5\)
Vậy \(Min_A=1\) khi \(1,5\le x\le2,5\)
Đặt \(A=\left|x-1,5\right|+\left|x-2,5\right|=\left|x-1,5\right|+\left|2,5-x\right|\)
Ta có: \(A\ge\left|x-1,5+2,5-x\right|=1\)
Dấu " = " xảy ra khi \(x-1,5\ge0;2,5-x\ge0\)
\(\Rightarrow x\ge1,5;x\le2,5\)
\(\Rightarrow1,5\le x\le2,5\)
Vậy \(MIN_A=1\) khi \(1,5\le x\le2,5\)