Đáp án:
i) \(x = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {x + 9} \right| = 2x\\
\to \left[ \begin{array}{l}
x + 9 = 2x\left( {DK:x \ge - 9} \right)\\
x + 9 = - 2x\left( {DK:x < - 9} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
3x = 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\left( {TM} \right)\\
x = 3\left( l \right)
\end{array} \right.\\
b)\left| {x + 6} \right| = 2x + 9\\
\to \left[ \begin{array}{l}
x + 6 = - 2x - 9\left( {DK:x < - 6} \right)\\
x + 6 = 2x + 9\left( {DK:x \ge - 6} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = - 15\\
x = - 3\left( {TM} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 3\\
x = - 5\left( l \right)
\end{array} \right.\\
c)\left| {2x - 3} \right| = 2x - 3\\
\to \left[ \begin{array}{l}
2x - 3 = 2x - 3\left( {ld} \right)\\
2x - 3 = - 2x + 3\left( {DK:x \le \dfrac{3}{2}} \right)
\end{array} \right.\\
\to 4x = 6\\
\to x = \dfrac{3}{2}\left( {TM} \right)\\
d)\left| {2x + 4} \right| = - 4x\\
\to \left[ \begin{array}{l}
2x + 4 = - 4x\left( {x \ge - 2} \right)\\
2x + 4 = 4x\left( {x < - 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
6x = - 4\\
2x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{2}{3}\left( {TM} \right)\\
x = 2\left( l \right)
\end{array} \right.\\
e)\left| {5x} \right| = 3x - 2\\
\to \left[ \begin{array}{l}
5x = 3x - 2\left( {DK:x \ge 0} \right)\\
5x = - 3x + 2\left( {DK:x < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - 2\\
8x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = \dfrac{1}{4}\left( l \right)
\end{array} \right.\\
\to x \in \emptyset \\
g)\left| { - \dfrac{5}{2}x} \right| = x - 12\\
\to \left[ \begin{array}{l}
- \dfrac{5}{2}x = x - 12\left( {DK:x \le 0} \right)\\
- \dfrac{5}{2}x = - x + 12\left( {DK:x > 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{7}{2}x = 12\\
\dfrac{3}{2}x = - 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{24}}{7}\left( l \right)\\
x = - 8\left( l \right)
\end{array} \right.\\
\to x \in \emptyset \\
h)\left| {5x} \right| = 3x + 2\\
\to \left[ \begin{array}{l}
5x = 3x + 2\left( {DK:x \ge 0} \right)\\
5x = - 3x - 2\left( {DK:x < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 2\\
8x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{4}\left( l \right)
\end{array} \right.\\
i)\left| { - 2x} \right| = 4x + 3\\
\to \left[ \begin{array}{l}
- 2x = 4x + 3\left( {DK:x \le 0} \right)\\
- 2x = - 4x - 3\left( {DK:x > 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
6x = - 3\\
2x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{2}\left( {TM} \right)\\
x = - \dfrac{3}{2}\left( l \right)
\end{array} \right.
\end{array}\)
