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The problem asks to express the following functions without absolute values. The three given functions are:

a. \( f(x) = |x| + |3x + 1| \)  
b. \( f(x) = 3 + |2x - 5| \)  
c. \( f(x) = 3|x - 2| - |x + 1| \)

To solve these, we need to break down each absolute value function based on its definition:
\[
|A| = 
\begin{cases} 
A & \text{if } A \geq 0 \\
-A & \text{if } A < 0
\end{cases}
\]
We'll apply this to each of the absolute values.

### a. \( f(x) = |x| + |3x + 1| \)

For \( |x| \):
- When \( x \geq 0 \), \( |x| = x \).
- When \( x < 0 \), \( |x| = -x \).

For \( |3x + 1| \):
- Solve \( 3x + 1 = 0 \), so \( x = -\frac{1}{3} \).
- When \( x \geq -\frac{1}{3} \), \( |3x + 1| = 3x + 1 \).
- When \( x < -\frac{1}{3} \), \( |3x + 1| = -(3x + 1) = -3x - 1 \).

Thus, we have three cases to consider:
1. For \( x \geq 0 \): \( f(x) = x + (3x + 1) = 4x + 1 \).
2. For \( -\frac{1}{3} \leq x < 0 \): \( f(x) = -x + (3x + 1) = 2x + 1 \).
3. For \( x < -\frac{1}{3} \): \( f(x) = -x + (-3x - 1) = -4x - 1 \).

### b. \( f(x) = 3 + |2x - 5| \)

Solve \( 2x - 5 = 0 \), so \( x = \frac{5}{2} = 2.5 \).

For \( |2x - 5| \):
- When \( x \geq 2.5 \), \( |2x - 5| = 2x - 5 \).
- When \( x < 2.5 \), \( |2x - 5| = -(2x - 5) = -2x + 5 \).

Thus, we have two cases:
1. For \( x \geq 2.5 \): \( f(x) = 3 + (2x - 5) = 2x - 2 \).
2. For \( x < 2.5 \): \( f(x) = 3 + (-2x + 5) = -2x + 8 \).

### c. \( f(x) = 3|x - 2| - |x + 1| \)

For \( |x - 2| \):
- Solve \( x - 2 = 0 \), so \( x = 2 \).
- When \( x \geq 2 \), \( |x - 2| = x - 2 \).
- When \( x < 2 \), \( |x - 2| = -(x - 2) = -x + 2 \).

For \( |x + 1| \):
- Solve \( x + 1 = 0 \), so \( x = -1 \).
- When \( x \geq -1 \), \( |x + 1| = x + 1 \).
- When \( x < -1 \), \( |x + 1| = -(x + 1) = -x - 1 \).

Thus, we have three cases to consider:
1. For \( x \geq 2 \): \( f(x) = 3(x - 2) - (x + 1) = 3x - 6 - x - 1 = 2x - 7 \).
2. For \( -1 \leq x < 2 \): \( f(x) = 3(-x + 2) - (x + 1) = -3x + 6 - x - 1 = -4x + 5 \).
3. For \( x < -1 \): \( f(x) = 3(-x + 2) - (-x - 1) = -3x + 6 + x + 1 = -2x + 7 \).

### Final Results:
- **a.**  
  \[
  f(x) = 
  \begin{cases} 
  4x + 1 & \text{if } x \geq 0 \\
  2x + 1 & \text{if } -\frac{1}{3} \leq x < 0 \\
  -4x - 1 & \text{if } x < -\frac{1}{3}
  \end{cases}
  \]

- **b.**  
  \[
  f(x) = 
  \begin{cases} 
  2x - 2 & \text{if } x \geq 2.5 \\
  -2x + 8 & \text{if } x < 2.5
  \end{cases}
  \]

- **c.**  
  \[
  f(x) = 
  \begin{cases} 
  2x - 7 & \text{if } x \geq 2 \\
  -4x + 5 & \text{if } -1 \leq x < 2 \\
  -2x + 7 & \text{if } x < -1
  \end{cases}
  \]

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. How do you determine the breakpoints for absolute value functions?
2. What are the different cases when solving absolute value equations?
3. How do absolute value functions relate to piecewise functions?
4. Can you graph the piecewise solutions for each function?
5. How do inequalities change when absolute values are involved?

**Tip**: Always solve for where the expressions inside absolute value are equal to zero. These points divide the domain into intervals where you can remove the absolute value.

Nyatakan fungsi berikut tidak dalam nilai mutlak: a. f(x) = |x| + |3x + 1| b. f(x) = 3 + |2x - 5| c. f(x) = 3|x - 2| - |x + 1|

This problem explores how to express functions without absolute values, focusing on solving f(x) = |x| + |3x + 1|, f(x) = 3 + |2x - 5|, and f(x) = 3|x - 2| - |x + 1|. Learn how to solve these using piecewise definitions of absolute values. Suitable for Grades 10-12 students.

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