Đáp án:
$\begin{array}{l}
a)C = - 2\left( {x - 1} \right) - 3\left| {x - 3} \right|\\
Khi:\left| {x - \frac{1}{2}} \right| = 1\\
\Leftrightarrow \left[ \begin{array}{l}
x - \frac{1}{2} = 1 \Leftrightarrow x = \frac{3}{2}\\
x - \frac{1}{2} = - 1 \Leftrightarrow x = \frac{{ - 1}}{2}
\end{array} \right.\\
+ Khi:x = \frac{3}{2}\\
\Leftrightarrow C = - 2.\left( {\frac{3}{2} - 1} \right) - 3.\left| {\frac{3}{2} - 3} \right|\\
= - 1 - 3.\frac{3}{2}\\
= \frac{{ - 11}}{2}\\
+ Khi:x = \frac{{ - 1}}{2}\\
\Leftrightarrow C = - 2.\left( { - \frac{1}{2} - 1} \right) - 3.\left| {\frac{{ - 1}}{2} - 3} \right|\\
= 6 - 3.\frac{7}{2}\\
= \frac{{ - 9}}{2}\\
b)C = - 2\left( {x - 1} \right) - 3\left| {x - 3} \right|\\
= - 2x + 2 - 3\left| {x - 3} \right|\\
= \left[ \begin{array}{l}
- 2x + 2 - 3\left( {x - 3} \right)\left( {khi:x \ge 3} \right)\\
- 2x + 2 - 3\left( {3 - x} \right)\left( {khi:x < 3} \right)
\end{array} \right.\\
= \left[ \begin{array}{l}
- 5x + 11\left( {khi:x \ge 3} \right)\\
x - 7\left( {khi:x < 3} \right)
\end{array} \right.\\
c)C = 2\\
\Leftrightarrow \left[ \begin{array}{l}
- 5x + 11 = 2\left( {khi:x \ge 3} \right)\\
x - 7 = 2\left( {khi:x < 3} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{9}{5}\left( {ktm} \right)\\
x = 9\left( {ktm} \right)
\end{array} \right.
\end{array}$
Vậy ko có x thỏa mãn để C=2
