Giải thích các bước giải:
\(\begin{array}{l}
a,\\
2\left| x \right| - \left| {x - 3} \right| = 8\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,x < 0 \Rightarrow \left\{ \begin{array}{l}
x < 0\\
x - 3 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 2.\left( { - x} \right) + \left( {x - 3} \right) = 8\\
\Leftrightarrow - x - 3 = 8\\
\Leftrightarrow x = - 11\,\,\,\,\,\,\left( {t/m} \right)\\
TH2:\,\,\,\,0 \le x \le 3 \Rightarrow \left\{ \begin{array}{l}
x \ge 0\\
x - 3 \le 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 2x + \left( {x - 3} \right) = 8\\
\Leftrightarrow 3x - 3 = 8\\
\Leftrightarrow x = \frac{{11}}{3}\,\,\,\,\,\,\,\left( L \right)\\
TH3:\,\,\,x > 3 \Rightarrow \left\{ \begin{array}{l}
x > 0\\
x - 3 > 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 2x - \left( {x - 3} \right) = 8\\
\Leftrightarrow x + 3 = 8\\
\Leftrightarrow x = 5\,\,\,\,\,\left( {t/m} \right)\\
\Rightarrow S = \left\{ { - 11;5} \right\}\\
b,\\
\left| {x + 1} \right| \le \left| x \right| - x + 2\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,x < - 1 \Rightarrow \left\{ \begin{array}{l}
x + 1 < 0\\
x < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x + 1} \right) \le - x - x + 2\\
\Leftrightarrow - x - 1 \le - 2x + 2\\
\Leftrightarrow x \le 3\\
\Rightarrow {S_1} = \left( { - \infty ; - 1} \right)\\
TH2:\,\,\, - 1 \le x \le 0 \Rightarrow \left\{ \begin{array}{l}
x + 1 \ge 0\\
x \le 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow x + 1 \le - x - x + 2\\
\Leftrightarrow 3x \le 1\\
\Leftrightarrow x \le \frac{1}{3}\\
\Rightarrow {S_2} = \left[ { - 1;0} \right]\\
TH3:\,\,\,x > 0 \Rightarrow \left\{ \begin{array}{l}
x + 1 > 0\\
x > 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow x + 1 \le x - x + 2\\
\Leftrightarrow x \le 1\\
\Rightarrow {S_3} = \left( {0;1} \right]\\
\Rightarrow S = {S_1} \cup {S_2} \cup {S_3} = \left( { - \infty ;1} \right]
\end{array}\)
