Đáp án:
$\begin{array}{l}
M = 3\left( {2x - 1} \right) - \left| {5 - 6x} \right|\\
= 6x - 3 - \left| {6x - 5} \right|\\
= \left[ \begin{array}{l}
6x - 3 - \left( {6x - 5} \right)\left( {khi:x \ge \dfrac{5}{6}} \right)\\
6x - 3 + 6x - 5\left( {khi:x < \dfrac{5}{6}} \right)
\end{array} \right.\\
= \left[ \begin{array}{l}
2\left( {khi:x \ge \dfrac{5}{6}} \right)\\
12x - 8\left( {khi:x < \dfrac{5}{6}} \right)
\end{array} \right.\\
N = 2\left| {2x + 1} \right| - \left| {x - 3} \right|\\
+ khi:x \ge 3\\
\Leftrightarrow N = 2\left( {2x + 1} \right) - \left( {x - 3} \right)\\
= 3x + 5\\
+ khi: - \dfrac{1}{2} \le x < 3\\
\Leftrightarrow N = 2\left( {2x + 1} \right) + \left( {x - 3} \right)\\
= 4x + 2 + x - 3\\
= 5x - 1\\
+ Khi:x < \dfrac{{ - 1}}{2}\\
\Leftrightarrow N = - 2\left( {2x + 1} \right) + x - 3\\
= - 4x - 2 + x - 3\\
= - 3x - 5\\
P = \left| {5x + 1} \right| + \left| {x + \dfrac{1}{3}} \right|\\
+ Khi:x \ge - \dfrac{1}{3}\\
\Leftrightarrow P = 5x + 1 + x + \dfrac{1}{3} = 6x + \dfrac{4}{3}\\
+ Khi: - \dfrac{1}{5} \le x < - \dfrac{1}{3}\\
\Leftrightarrow P = 5x + 1 - x - \dfrac{1}{3} = 4x + \dfrac{2}{3}\\
+ Khi:x < \dfrac{{ - 1}}{5}\\
\Leftrightarrow P = - 5x - 1 - x - \dfrac{1}{3} = - 6x - \dfrac{4}{3}
\end{array}$
