Đáp án:
b. \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{{15}}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {4x - 4} \right| + \left| {x - 2} \right| = 5\\
\to \left[ \begin{array}{l}
4x - 4 + x - 2 = 5\left( {DK:x \ge 2} \right)\\
4x - 4 - x + 2 = 5\left( {DK:2 > x \ge 1} \right)\\
4x - 4 + x - 2 = - 5\left( {DK:1 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = 11\\
3x = 7\\
5x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{11}}{5}\left( {TM} \right)\\
x = \dfrac{7}{3}\left( l \right)\\
x = \dfrac{1}{5}\left( {TM} \right)
\end{array} \right.\\
b.\left| {x + 3} \right| + \left| {x + 4} \right| = 8\\
\to \left[ \begin{array}{l}
x + 3 + x + 4 = 8\left( {DK:x \ge - 3} \right)\\
- x - 3 + x + 4 = 8\left( {DK: - 3 > x \ge - 4} \right)\\
x + 3 + x + 4 = - 8\left( {DK: - 4 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 1\\
1 = 8\left( l \right)\\
2x = - 15
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{{15}}{2}
\end{array} \right.\left( {TM} \right)
\end{array}\)
