Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\left| {x - 1} \right| + \left| {2x + 1} \right| = \left| {3x} \right|\\
 \Leftrightarrow {\left( {\left| {x - 1} \right| + \left| {2x - 1} \right|} \right)^2} = {\left| {3x} \right|^2}\\
 \Leftrightarrow {\left( {x - 1} \right)^2} + 2.\left| {x - 1} \right|.\left| {2x - 1} \right| + {\left( {2x - 1} \right)^2} = 9{x^2}\\
 \Leftrightarrow \left( {{x^2} - 2x + 1} \right) + 2.\left| {\left( {x - 1} \right)\left( {2x - 1} \right)} \right| + \left( {4{x^2} - 4x + 1} \right) = 9{x^2}\\
 \Leftrightarrow \left( {5{x^2} - 6x + 2} \right) + 2.\left| {2{x^2} - 3x + 1} \right| = 9{x^2}\\
 \Leftrightarrow 2.\left| {2{x^2} - 3x + 1} \right| = 4{x^2} + 6x - 2\\
 \Leftrightarrow \left| {2{x^2} - 3x + 1} \right| = 2{x^2} + 3x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} + 3x - 1 \ge 0\\
\left[ \begin{array}{l}
2{x^2} - 3x + 1 = 2{x^2} + 3x - 1\\
2{x^2} - 3x + 1 =  - 2{x^2} - 3x + 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} + 3x - 1 \ge 0\\
\left[ \begin{array}{l}
6x = 2\\
4{x^2} = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} + 3x - 1 \ge 0\\
\left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{1}{3}\\
b,\\
\left| {3x - 2} \right| = {x^2} + 2x + 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 2x + 3 \ge 0\\
\left[ \begin{array}{l}
3x - 2 = {x^2} + 2x + 3\\
3x - 2 =  - {x^2} - 2x - 3
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{x^2} + 2x + 1} \right) + 2 \ge 0\\
\left[ \begin{array}{l}
{x^2} - x + 5 = 0\\
{x^2} + 5x + 1 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} + 2 \ge 0,\,\,\,\forall x\\
x = \dfrac{{ - 5 \pm \sqrt {21} }}{2}
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{{ - 5 \pm \sqrt {21} }}{2}\\
c,\\
\left| {{x^2} + 3x - 2} \right| = x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
\left[ \begin{array}{l}
{x^2} + 3x - 2 = x - 1\\
{x^2} + 3x - 2 =  - x + 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
{x^2} + 2x - 1 = 0\\
{x^2} + 4x - 3 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x =  - 1 \pm \sqrt 2 \\
x =  - 2 \pm \sqrt 7 
\end{array} \right.
\end{array} \right.\\
 \Rightarrow ptvn\\
d,\\
\left| {{x^3} - 1} \right| = \left| {{x^2} - 3x + 2} \right|\\
 \Leftrightarrow \left| {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} \right| = \left| {\left( {x - 1} \right)\left( {x - 2} \right)} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
\left| {{x^2} + x + 1} \right| = \left| {x - 2} \right|
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
{x^2} + x + 1 = x - 2\\
{x^2} + x + 1 =  - x + 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
{x^2} =  - 3\,\,\,\,\left( L \right)\\
{x^2} + 2x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 1 \pm \sqrt 2 
\end{array} \right.
\end{array}\)