a) Ta có: 

VT = |x + 1| + |x + 2| + |2x - 3| \(\ge\)|x + 1 + x + 2| + |3 - 2x| =  |2x + 3| + |3 - 2x| \(\ge\)|2x + 3 + 3 - 2x| = 6

VP = 6

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\)  => \(\orbr{\begin{cases}x\ge-1\\x\le-2\end{cases}}\)và \(-\frac{3}{2}\le x\le\frac{3}{2}\)

<=> \(-1\le x\le\frac{3}{2}\)

b) Ta có: VT = |x + 1| + |x - 2| + |x - 3| + |x - 5| = (|x + 1| + |5 - x|) + (|x - 2| + |3 - x|) \(\ge\)|x + 1 + 5 - x| + |x - 2 + 3 - x| = |6| + |1| = 7

VP = 7

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(5-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{cases}}\) <=> \(\hept{\begin{cases}-1\le x\le5\\2\le x\le3\end{cases}}\) <=> \(2\le x\le3\)

a. ta có :

\(\hept{\begin{cases}\left|x-1\right|+\left|x-4\right|\ge\left|x-1-x+4\right|=3\\\left|x-2\right|+\left|x-3\right|\ge\left|x-2-x+3\right|=1\\\left|2x-5\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\)thay lại thấy thỏa mãn . Vậy x=5/2 là nghiệm

b.ta có 

\(\hept{\begin{cases}\left|x+1\right|+\left|x-1\right|\ge\left|x+1-x+1\right|=2\\\left|x+2\right|+\left|x-5\right|\ge\left|x+2-x+5\right|=7\\\left|3x+2\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\)thay lại thấy thỏa mãn . Vậy x=-2/3 là nghiệm

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).

Ta có bất đẳng thức giá trị tuyệt đối: 

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Dấu \(=\)khi \(AB\ge0\).

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

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

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

\(\ge\left|2x+3+3-2x\right|=6\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).

e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)

\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)

\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)

\(=4+7=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).

Do đó phương trình đã cho vô nghiệm.