Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\left| {x + 1} \right| = \left| {{x^2} - 3x + 2} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = {x^2} - 3x + 2\\
x + 1 =  - \left( {{x^2} - 3x + 2} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left( {{x^2} - 3x + 2} \right) - \left( {x + 1} \right) = 0\\
\left( {{x^2} - 3x + 2} \right) + \left( {x + 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4x + 1 = 0\\
{x^2} - 2x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow x = 2 \pm \sqrt 3 \\
b,\\
\left| {{x^2} - x} \right| = 2x - 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - 3 \ge 0\\
\left[ \begin{array}{l}
{x^2} - x = 2x - 3\\
{x^2} - x =  - \left( {2x - 3} \right)
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
\left[ \begin{array}{l}
{x^2} - x - \left( {2x - 3} \right) = 0\\
{x^2} - x + \left( {2x - 3} \right) = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
{x^2} - 3x + 3 = 0\\
{x^2} + x - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x = \dfrac{{ - 1 \pm \sqrt {13} }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {ptvn} \right)\\
c,\\
DKXD:\,\,\,{x^2} - 2 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 2 \\
x \le  - \sqrt 2 
\end{array} \right.\\
\sqrt {{x^2} - 2}  = x + 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 1 \ge 0\\
{x^2} - 2 = {\left( {x + 1} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
{x^2} - 2 = {x^2} + 2x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
2x =  - 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
x =  - \dfrac{3}{2}
\end{array} \right.\,\,\,\,\,\,\left( {ptvn} \right)\\
d,\\
DKXD:\,\,\,\,4x - 1 \ge 0 \Leftrightarrow x \ge \dfrac{1}{4}\\
\sqrt {4x - 1}  = {x^2} + 2x\\
 \Leftrightarrow 2\sqrt {4x - 1}  = 2{x^2} + 4x\\
 \Leftrightarrow 2{x^2} + 4x - 2\sqrt {4x - 1}  = 0\\
 \Leftrightarrow 2{x^2} + \left[ {\left( {4x - 1} \right) - 2\sqrt {4x - 1}  + 1} \right] = 0\\
 \Leftrightarrow 2{x^2} + {\left( {\sqrt {4x - 1}  - 1} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 0\\
{\left( {\sqrt {4x - 1}  - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
\sqrt {4x - 1}  = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}
\end{array} \right.\,\,\,\,\,\,\left( {vn} \right)
\end{array}\)