Giải thích các bước giải:

 +) Nếu x>2 thì 

$\begin{array}{l} \left\{ \begin{array}{l} {x^2} - x > 0\\ 2x - 4 > 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} - x} \right| = {x^2} - x\\ \left| {2x - 4} \right| = 2x - 4 \end{array} \right. \end{array}$

Khi đó pt trở thành:

$\begin{array}{l} {x^2} - x + 2x - 4 = 3\\  \Leftrightarrow {x^2} + x - 7 = 0\\  \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{{29}}{4}\\  \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt {29} }}{2}\\ x + \frac{1}{2} = \frac{{ - \sqrt {29} }}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt {29}  - 1}}{2}\\ x = \frac{{ - \sqrt {29}  - 1}}{2} \end{array} \right.\\  \Leftrightarrow x = \frac{{\sqrt {29}  - 1}}{2} \end{array}$

+) Nếu 1≤x≤2 thì 

$\begin{array}{l} \left\{ \begin{array}{l} {x^2} - x ≥ 0\\ 2x - 4 ≤ 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} - x} \right| = {x^2} - x\\ \left| {2x - 4} \right| = -2x + 4 \end{array} \right. \end{array}$

Khi đó pt trở thành:

$\begin{array}{l} {x^2} - x - 2x + 4 = 3\\  \Leftrightarrow {x^2} - 3x + 1 = 0\\  \Leftrightarrow {(x - \frac{3}{2})^2} = \frac{5}{4}\\  \Leftrightarrow \left[ \begin{array}{l} x - \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x - \frac{3}{2} = \frac{{ - \sqrt 5 }}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5  + 3}}{2}\\ x = \frac{{ - \sqrt 5  + 3}}{2} \end{array} \right. \end{array}$

(không thoả mãn)

+) Nếu 0≤x≤1 thì 

$\begin{array}{l} \left\{ \begin{array}{l} {x^2} - x ≤ 0\\ 2x - 4 ≤ 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} - x} \right| = -{x^2} + x\\ \left| {2x - 4} \right| = -2x + 4 \end{array} \right. \end{array}$

Khi đó pt trở thành:

$\begin{array}{l}  - {x^2} + x - 2x + 4 = 3\\  \Leftrightarrow x + x - 1 = 0\\  \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{5}{4}\\  \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt 5 }}{2}\\ x + \frac{1}{2} = \frac{{ - \sqrt 5 }}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5  - 1}}{2}\\ x = \frac{{ - \sqrt 5  - 1}}{2} \end{array} \right.\\  \Leftrightarrow x = \frac{{\sqrt 5  - 1}}{2} \end{array}$

+) Nếu x<0 thì 

$\begin{array}{l} \left\{ \begin{array}{l} {x^2} - x > 0\\ 2x - 4 < 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} - x} \right| = {x^2} - x\\ \left| {2x - 4} \right| = 2x - 4 \end{array} \right. \end{array}$

Khi đó pt trở thành:

$\begin{array}{l} {x^2} - x - 2x + 4 = 3\\  \Leftrightarrow x - 3x + 1 = 0\\  \Leftrightarrow {(x - \frac{3}{2})^2} = \frac{5}{4}\\  \Leftrightarrow \left[ \begin{array}{l} x - \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x - \frac{3}{2} = \frac{{ - \sqrt 5 }}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5  + 3}}{2}\\ x = \frac{{ - \sqrt 5  + 3}}{2} \end{array} \right. \end{array}$

(không thoả mãn)