Đáp án:
$\begin{array}{l}
1)\left| { - {x^2} + x - 1} \right| \le 2x + 5\left( {dk:x \ge - \frac{5}{2}} \right)\\
\Rightarrow \left[ \begin{array}{l}
- {x^2} + x - 1 \le 2x + 5\\
- {x^2} + x - 1 \ge - 2x - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + x + 6 \ge 0\left( {luon\,dung} \right)\\
{x^2} - 3x - 4 \le 0
\end{array} \right.\\
\Rightarrow \left( {x - 4} \right)\left( {x + 1} \right) \le 0\\
\Rightarrow - 1 \le x \le 4\\
Vay\, - 1 \le x \le 4\\
2)\left| {4 - 3x} \right| \le 8\\
\Rightarrow 9{x^2} - 24x + 16 \le 64\\
\Rightarrow 9{x^2} - 24x - 48 \le 0\\
\Rightarrow - \frac{4}{3} \le x \le 4\\
3)\left| {2x - 3} \right| \le x + 12\left( {x \ge - 12} \right)\\
\Rightarrow 4{x^2} - 12x + 9 \le {x^2} + 24x + 144\\
\Rightarrow 3{x^2} - 36x - 135 \le 0\\
\Rightarrow - 3 \le x \le 15\\
4)\left| {\frac{{2x - 1}}{{x - 1}}} \right| > 2\left( {x \ne 1} \right)\\
\Rightarrow 4{x^2} - 4x + 1 > 2\left( {{x^2} - 2x + 1} \right)\\
\Rightarrow 2{x^2} > 1\\
\Rightarrow {x^2} > \frac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x > \frac{{\sqrt 2 }}{2};x \ne 1\\
x < - \frac{{\sqrt 2 }}{2}
\end{array} \right.\\
5)\left| {x - 15} \right| \ge 3\\
\Rightarrow {x^2} - 30x + 225 \ge 9\\
\Rightarrow {x^2} - 30x + 216 \ge 0\\
\Rightarrow 12 \le x \le 18\\
6)\left| {x - 2} \right| > x + 1\\
\Rightarrow {x^2} - 4x + 4 > {x^2} + 2x + 1\\
\Rightarrow 6x < 3\\
\Rightarrow x < \frac{1}{2}
\end{array}$
