Bài 1: 

a) Ta có: \(4\left(x-5\right)+2x=-13\)

\(\Leftrightarrow6x-20=-13\)

\(\Leftrightarrow6x=7\)

hay \(x=\dfrac{7}{6}\)

b) Ta có: \(\dfrac{2x-1}{4}-3=\dfrac{1+x}{6}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{12}-\dfrac{36}{12}=\dfrac{2\left(x+1\right)}{12}\)

Suy ra: \(6x-3-36=2x+2\)

\(\Leftrightarrow4x=41\)

hay \(x=\dfrac{41}{4}\)

c) ĐKXĐ: \(x\notin\left\{0;3\right\}\)

Ta có: \(\dfrac{x+3}{x-3}-\dfrac{8}{x}=\dfrac{18}{x^2-3x}\)

\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}-\dfrac{8\left(x-3\right)}{x\left(x-3\right)}=\dfrac{18}{x\left(x-3\right)}\)

Suy ra: \(x^2+3x-8x+24-18=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={2}

Bài 2: 

a) Ta có: 3x-11>5-x

\(\Leftrightarrow3x+x>5+11\)

\(\Leftrightarrow4x>16\)

hay x>4

Vậy: S={x|x>4}

b) Ta có: \(\dfrac{2x+1}{3}-\dfrac{3x-2}{6}\ge x+\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{6}-\dfrac{3x-2}{6}\ge\dfrac{6x}{6}+\dfrac{1}{6}\)

Suy ra: \(4x+2-3x+2-6x-1\ge0\)

\(\Leftrightarrow-5x\ge-3\)

hay \(x< \dfrac{3}{5}\)

Vậy: S={x|\(x< \dfrac{3}{5}\)}