Đáp án:

$\begin{array}{l}
c)\dfrac{{x + 1}}{3} > \dfrac{{2x - 1}}{6} - 2\\
 \Leftrightarrow \dfrac{{2\left( {x + 1} \right)}}{6} > \dfrac{{2x - 1 - 12}}{6}\\
 \Leftrightarrow 2x + 2 > 2x - 13\\
 \Leftrightarrow 2 >  - 13\left( {tm} \right)\\
Vậy\,x \in R\\
d)\dfrac{{2x + 1}}{3} > \dfrac{{2x - 1}}{6} - 2\\
 \Leftrightarrow \dfrac{{2\left( {2x + 1} \right)}}{6} > \dfrac{{2x - 1 - 12}}{6}\\
 \Leftrightarrow 4x + 2 > 2x - 13\\
 \Leftrightarrow 2x >  - 15\\
 \Leftrightarrow x > \dfrac{{ - 15}}{2}\\
Vậy\,x > \dfrac{{ - 15}}{2}\\
e)\dfrac{{x + 5}}{6} - \dfrac{{2x + 1}}{3} \le \dfrac{{x + 3}}{2}\\
 \Leftrightarrow \dfrac{{x + 5 - 2\left( {2x + 1} \right)}}{6} \le \dfrac{{3\left( {x + 3} \right)}}{6}\\
 \Leftrightarrow x + 5 - 4x - 2 \le 3x + 9\\
 \Leftrightarrow 6x \ge  - 12\\
 \Leftrightarrow x \ge  - 2\\
Vậy\,x \ge  - 2\\
f)\dfrac{{5x + 4}}{6} - \dfrac{{2x - 1}}{{12}} \ge 4\\
 \Leftrightarrow \dfrac{{2\left( {5x + 4} \right) - 2x + 1}}{{12}} \ge \dfrac{{4.12}}{{12}}\\
 \Leftrightarrow 10x + 8 - 2x + 1 \ge 48\\
 \Leftrightarrow 8x \ge 39\\
 \Leftrightarrow x \ge \dfrac{{39}}{8}\\
Vậy\,x \ge \dfrac{{39}}{8}
\end{array}$

Đáp án:

`c. \frac{x+1}{3}>\frac{2x-1}{6}-2`

`<=> 2(x+1)>2x-1-12`

`<=> 2x+2>2x-1-12`

`<=> 2> -1-12`

`<=> 2> -13` (luôn đúng)

`=> x\inRR`

Vậy `S=RR`

`d. \frac{2x+1}{3}>\frac{2x-1}{6}-2`

`<=> 2(2x+1)>2x-1-12`

`<=> 4x+2>2x-1-12`

`<=> 2x> -15`

`<=> x> -15/2`

Vậy `S={x|x>-15/2}`

`e. \frac{x+5}{6}-\frac{2x+1}{3}\lefrac{x+3}{2}`

`<=> x+5-2(2x+1)\le3(x+3)`

`<=> x+5-4x-2\le3x+9`

`<=> -3x+3\le3x+9`

`<=> -6x\le6`

`<=> x\ge-1`

Vậy `S={x|x\ge-1}`

`f. \frac{5x+4}{6}-\frac{2x-1}{12}\ge4`

`<=> 2(5x+4)-2x+1\ge48`

`<=> 10x+8-2x+1\ge48`

`<=> 8x+9\ge48`

`<=> 8x\ge39`

`<=> x\ge39/8`

Vậy `S={x|x\ge39/8}`