Đáp án:
c) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {2x - 1} \right| = \left| {2x - 5} \right|\\
\to \left[ \begin{array}{l}
2x - 1 = 2x - 5\left( l \right)\\
2x - 1 = - 2x + 5
\end{array} \right.\\
\to 4x = 6\\
\to x = \dfrac{3}{2}\\
b)\left| {7 - x} \right| = \left| {2x - 3} \right|\\
\to \left[ \begin{array}{l}
7 - x = 2x - 3\left( {DK:x \le 7} \right)\\
- 7 + x = 2x - 3\left( {DK:x > 7} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 10\\
x = - 4\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{10}}{3}\\
c)\left| {x - 4} \right| + \left| {{x^2} - 5x + 4} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 4 = 0\\
{x^2} - 5x + 4 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
\left( {x - 1} \right)\left( {x - 4} \right) = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
x = 1\\
x = 4
\end{array} \right.\\
KL:x = 4\\
d)\left| {\dfrac{{{x^2} - x - 2}}{{x + 1}}} \right| - \left| x \right| = 0\\
\to \left| {\dfrac{{{x^2} - x - 2}}{{x + 1}}} \right| = \left| x \right|\\
\to \left[ \begin{array}{l}
\dfrac{{{x^2} - x - 2}}{{x + 1}} = x\left( {DK:x \ge 0} \right)\\
\dfrac{{{x^2} - x - 2}}{{x + 1}} = - x\left( {DK:x < 0;x \ne - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - x - 2 = {x^2} + x\\
{x^2} - x - 2 = - {x^2} - x
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - 2\\
2{x^2} = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = 1\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
