Ta có: 

\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\ge\left|x+2+1-x\right|=3\)

Dấu \(=\)khi \(\left(x+2\right)\left(1-x\right)\ge0\Leftrightarrow-2\le x\le1\).

Do đó \(\left|x+2\right|+\left|3x-1\right|+\left|x-1\right|=3\)

\(\Leftrightarrow\hept{\begin{cases}-2\le x\le1\\3x-1=0\end{cases}}\Leftrightarrow x=\frac{1}{3}\).

a, TH1 : \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}\Leftrightarrow}x>\frac{1}{3}}\)

TH2 : \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}\Leftrightarrow x< -\frac{2}{5}}\)

Vậy x > 1/3 ; x < -2/5 

b, Vì \(x+1>x+\frac{3}{5}\)

nên \(\hept{\begin{cases}x+1>0\\x+\frac{3}{5}< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -\frac{3}{5}\end{cases}\Leftrightarrow-1< x< -\frac{3}{5}}}\)

\(\left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right)>0\)

\(\orbr{\begin{cases}x>\frac{1}{3}\\x< -\frac{2}{5}\end{cases}}\)

<cùng dương, cùng âm>

\(A=\left|x-2002\right|+\left|x-2003\right|=\left|x-2002\right|+\left|2003-x\right|\ge\left|-2002+2003\right|=1\)

Dấu ''='' xảy ra khi \(\left(x-2002\right)\left(2003-x\right)\ge0\Leftrightarrow2002\le x\le2003\)

Vậy GTNN của A bằng 1 tại 2002 =< x =< 2003 

\(B=5,5-\left|2x-5\right|\le5,5\)

Dấu ''='' xảy ra khi x = 5/2

Vậy GTLN của B bằng 5,5 tại x = 5/2 

Vì \(\left(2x-1\right)^{2008}\ge0\forall x;\left(y-\frac{2}{5}\right)^{2008}\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)

\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\forall x;y;z\)

mà \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

Đẳng thức xảy ra khi \(x=\frac{1}{2};y=\frac{2}{5};z=-\frac{9}{10}\)

Vì \(\hept{\begin{cases}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\frac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y+z\right|\ge0\forall x,y,z\end{cases}}\)

\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\\frac{1}{2}+\frac{2}{5}+z=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=-\frac{9}{10}\end{cases}}\)

Vậy \(\left(x,y,z\right)=\left(\frac{1}{2};\frac{2}{5};-\frac{9}{10}\right)\)