Let $ c$ be a constant number. Find the real value of $ x$ such that $ \int_0^c |t\minus{}x|\ dt\equal{}\int_0^x |t\minus{}c|\ dt$.
To solve the problem, we need to find the real value of \( x \) such that the following integral equation holds:
\[
\int_0^c |t - x| \, dt = \int_0^x |t - c| \, dt
\]

We will consider different cases based on the relative values of \( x \) and \( c \).

### Case 1: \( x \geq c \geq 0 \)

1. **Evaluate \(\int_0^c |t - x| \, dt\):**
   Since \( x \geq c \geq 0 \), \( t - x \leq 0 \) for \( t \in [0, c] \). Therefore, \( |t - x| = x - t \).
   \[
   \int_0^c |t - x| \, dt = \int_0^c (x - t) \, dt = \left[ xt - \frac{t^2}{2} \right]_0^c = xc - \frac{c^2}{2}
   \]

2. **Evaluate \(\int_0^x |t - c| \, dt\):**
   Since \( x \geq c \geq 0 \), we split the integral into two parts:
   \[
   \int_0^x |t - c| \, dt = \int_0^c |t - c| \, dt + \int_c^x |t - c| \, dt
   \]
   For \( t \in [0, c] \), \( |t - c| = c - t \):
   \[
   \int_0^c (c - t) \, dt = \left[ ct - \frac{t^2}{2} \right]_0^c = c^2 - \frac{c^2}{2} = \frac{c^2}{2}
   \]
   For \( t \in [c, x] \), \( |t - c| = t - c \):
   \[
   \int_c^x (t - c) \, dt = \left[ \frac{t^2}{2} - ct \right]_c^x = \left( \frac{x^2}{2} - cx \right) - \left( \frac{c^2}{2} - c^2 \right) = \frac{x^2}{2} - cx + \frac{c^2}{2}
   \]
   Combining these results:
   \[
   \int_0^x |t - c| \, dt = \frac{c^2}{2} + \left( \frac{x^2}{2} - cx + \frac{c^2}{2} \right) = c^2 - cx + \frac{x^2}{2}
   \]

3. **Set the integrals equal and solve for \( x \):**
   \[
   xc - \frac{c^2}{2} = c^2 - cx + \frac{x^2}{2}
   \]
   Rearrange and simplify:
   \[
   0 = \frac{x^2}{2} - 2cx + \frac{3c^2}{2}
   \]
   Multiply through by 2:
   \[
   0 = x^2 - 4cx + 3c^2
   \]
   Solve the quadratic equation:
   \[
   x = \frac{4c \pm \sqrt{16c^2 - 12c^2}}{2} = \frac{4c \pm 2c}{2} = 2c \pm c
   \]
   Thus, \( x = 3c \) or \( x = c \).

### Case 2: \( c \geq x \geq 0 \)

1. **Evaluate \(\int_0^c |t - x| \, dt\):**
   Since \( c \geq x \geq 0 \), we split the integral into two parts:
   \[
   \int_0^c |t - x| \, dt = \int_0^x |t - x| \, dt + \int_x^c |t - x| \, dt
   \]
   For \( t \in [0, x] \), \( |t - x| = x - t \):
   \[
   \int_0^x (x - t) \, dt = \left[ xt - \frac{t^2}{2} \right]_0^x = x^2 - \frac{x^2}{2} = \frac{x^2}{2}
   \]
   For \( t \in [x, c] \), \( |t - x| = t - x \):
   \[
   \int_x^c (t - x) \, dt = \left[ \frac{t^2}{2} - xt \right]_x^c = \left( \frac{c^2}{2} - xc \right) - \left( \frac{x^2}{2} - x^2 \right) = \frac{c^2}{2} - xc + \frac{x^2}{2}
   \]
   Combining these results:
   \[
   \int_0^c |t - x| \, dt = \frac{x^2}{2} + \left( \frac{c^2}{2} - xc + \frac{x^2}{2} \right) = \frac{c^2}{2} - xc + x^2
   \]

2. **Evaluate \(\int_0^x |t - c| \, dt\):**
   Since \( c \geq x \geq 0 \), \( t - c \leq 0 \) for \( t \in [0, x] \). Therefore, \( |t - c| = c - t \).
   \[
   \int_0^x |t - c| \, dt = \int_0^x (c - t) \, dt = \left[ ct - \frac{t^2}{2} \right]_0^x = cx - \frac{x^2}{2}
   \]

3. **Set the integrals equal and solve for \( x \):**
   \[
   \frac{c^2}{2} - xc + x^2 = cx - \frac{x^2}{2}
   \]
   Rearrange and simplify:
   \[
   0 = x^2 - 4cx + c^2
   \]
   Solve the quadratic equation:
   \[
   x = \frac{4c \pm \sqrt{16c^2 - 12c^2}}{2} = \frac{4c \pm 2c}{2} = 2c \pm c
   \]
   Thus, \( x = \frac{c}{3} \) or \( x = c \).

### Case 3: \( x \leq c \leq 0 \)

1. **Evaluate \(\int_0^c |t - x| \, dt\):**
   Since \( x \leq c \leq 0 \), \( t - x \geq 0 \) for \( t \in [0, c] \). Therefore, \( |t - x| = t - x \).
   \[
   \int_0^c |t - x| \, dt = \int_0^c (t - x) \, dt = \left[ \frac{t^2}{2} - xt \right]_0^c = \frac{c^2}{2} - cx
   \]

2. **Evaluate \(\int_0^x |t - c| \, dt\):**
   Since \( x \leq c \leq 0 \), we split the integral into two parts:
   \[
   \int_0^x |t - c| \, dt = \int_0^c |t - c| \, dt + \int_c^x |t - c| \, dt
   \]
   For \( t \in [0, c] \), \( |t - c| = c - t \):
   \[
   \int_0^c (c - t) \, dt = \left[ ct - \frac{t^2}{2} \right]_0^c = c^2 - \frac{c^2}{2} = \frac{c^2}{2}
   \]
   For \( t \in [c, x] \), \( |t - c| = t - c \):
   \[
   \int_c^x (t - c) \, dt = \left[ \frac{t^2}{2} - ct \right]_c^x = \left( \frac{x^2}{2} - cx \right) - \left( \frac{c^2}{2} - c^2 \right) = \frac{x^2}{2} - cx + \frac{c^2}{2}
   \]
   Combining these results:
   \[
   \int_0^x |t - c| \, dt = \frac{c^2}{2} + \left( \frac{x^2}{2} - cx + \frac{c^2}{2} \right) = c^2 - cx + \frac{x^2}{2}
   \]

3. **Set the integrals equal and solve for \( x \):**
   \[
   \frac{c^2}{2} - cx = cx - \frac{x^2}{2}
   \]
   Rearrange and simplify:
   \[
   0 = x^2 - 4cx + c^2
   \]
   Solve the quadratic equation:
   \[
   x = \frac{4c \pm \sqrt{16c^2 - 12c^2}}{2} = \frac{4c \pm 2c}{2} = 2c \pm c
   \]
   Thus, \( x = \frac{c}{3} \) or \( x = c \).

### Case 4: \( xc < 0 \)

1. **Let \( c > 0 \) and \( x < 0 \):**
   \[
   \int_0^c |t - x| \, dt = \int_0^c (t - x) \, dt = \frac{c^2}{2} - cx
   \]
   \[
   \int_0^x |t - c| \, dt = \int_0^x (c - t) \, dt = cx - \frac{x^2}{2}
   \]
   Set the integrals equal and solve for \( x \):
   \[
   \frac{c^2}{2} - cx = cx - \frac{x^2}{2}
   \]
   Rearrange and simplify:
   \[
   0 = x^2 - 4cx + c^2
   \]
   Solve the quadratic equation:
   \[
   x = \frac{4c \pm \sqrt{16c^2 - 12c^2}}{2} = \frac{4c \pm 2c}{2} = 2c \pm c
   \]
   Thus, \( x = (2 \pm \sqrt{3})c \), which is not possible since \( x < 0 \).

2. **By symmetry, the same situation arises when \( x > 0 \) and \( c < 0 \).**

Therefore, the complete set of solutions is:
\[
x = \left\{ \frac{c}{3}, c, 3c \right\}
\]

The final answer is \( \boxed{ x = \left\{ \frac{c}{3}, c, 3c \right\} } \)