BĐT : |a|+|b| \(\ge\) |a+b| ,dấu "=" xảy ra <=> a.b \(\ge\) 0
Có: \(A=\left|x-1\right|+\left|x-2\right|=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\) (vì |x-2|=|2-x|)
Dấu "=" xảy ra <=> (x-1)(2-x) \(\ge\) 0 <=> 1 \(\le\) x \(\le\) 2
Vậy................