\(\left|x-1\right|+\left|x-2\right|=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)

Dau "=" xay ra <=> \(\left(x-1\right)\left(2-x\right)\ge0\Rightarrow1\le x\le2\)

Áp dụng BĐT |a|+|b|>=|a+b|

Ta có:

\(\left|x+3\right|+\left|x-2\right|\ge\left|x+3+2-x\right|=5\)

\(\Rightarrow H\ge5\)

Dấu "=" xảy ra khi \(x+3=0\Leftrightarrow x=-3;x-2=0\Leftrightarrow x=2\)

Vậy MinH=5<=>x=-3 hoặc x=2

Ta có : A = |x - 2001| + |x - 1|

               =  |x - 2001| + |1- x|

             \(\ge\) |x - 2001 + 1 - x|

               = 2000 

Dấu "=" xảy ra <=> \(\left(1-x\right)\left(x-2001\right)\ge0\)  

=> \(\hept{\begin{cases}1-x\ge0\\x-2001\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\le1\\x\ge2001\end{cases}\Rightarrow}x\in\varnothing}\)

hoặc \(\hept{\begin{cases}1-x\le0\\x-2001\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge1\\x\le2001\end{cases}\Rightarrow}1\le x\le2001}\)

Vậy MIN A = 2000 <=>  \(1\le x\le2001\)