Assume $c$ is a real number. If there exists $x\in[1,2]$ such that $\max\left\{\left |x+\frac cx\right |, \left |x+\frac cx + 2\right |\right\}\geq 5$, please find the value range of $c$.
1. Let \( t = x + \frac{c}{x} \). The given condition is:
   \[
   \max\left\{\left| t \right|, \left| t + 2 \right|\right\} \geq 5
   \]
   This means that either \( \left| t \right| \geq 5 \) or \( \left| t + 2 \right| \geq 5 \).

2. We need to analyze both cases:
   - Case 1: \( \left| t \right| \geq 5 \)
   - Case 2: \( \left| t + 2 \right| \geq 5 \)

3. For Case 1:
   \[
   \left| t \right| \geq 5 \implies t \leq -5 \text{ or } t \geq 5
   \]

4. For Case 2:
   \[
   \left| t + 2 \right| \geq 5 \implies t + 2 \leq -5 \text{ or } t + 2 \geq 5
   \]
   Simplifying, we get:
   \[
   t \leq -7 \text{ or } t \geq 3
   \]

5. Combining both cases, we have:
   \[
   t \leq -7 \text{ or } t \geq 5 \text{ or } t \geq 3
   \]
   This simplifies to:
   \[
   t \leq -7 \text{ or } t \geq 3
   \]

6. Now, substituting back \( t = x + \frac{c}{x} \):
   \[
   x + \frac{c}{x} \leq -7 \text{ or } x + \frac{c}{x} \geq 3
   \]

7. We need to find the range of \( c \) for \( x \in [1, 2] \).

8. For \( x + \frac{c}{x} \leq -7 \):
   \[
   x + \frac{c}{x} \leq -7 \implies \frac{c}{x} \leq -7 - x \implies c \leq -x^2 - 7x
   \]

9. For \( x + \frac{c}{x} \geq 3 \):
   \[
   x + \frac{c}{x} \geq 3 \implies \frac{c}{x} \geq 3 - x \implies c \geq -x^2 + 3x
   \]

10. We need to find the minimum and maximum values of the expressions \( -x^2 - 7x \) and \( -x^2 + 3x \) for \( x \in [1, 2] \).

11. For \( -x^2 - 7x \):
    \[
    f(x) = -x^2 - 7x
    \]
    The function is a downward parabola. Evaluating at the endpoints:
    \[
    f(1) = -1 - 7 = -8
    \]
    \[
    f(2) = -4 - 14 = -18
    \]
    Thus, the minimum value is:
    \[
    \min_{x \in [1, 2]} (-x^2 - 7x) = -18
    \]

12. For \( -x^2 + 3x \):
    \[
    g(x) = -x^2 + 3x
    \]
    The function is a downward parabola. Evaluating at the endpoints:
    \[
    g(1) = -1 + 3 = 2
    \]
    \[
    g(2) = -4 + 6 = 2
    \]
    Thus, the maximum value is:
    \[
    \max_{x \in [1, 2]} (-x^2 + 3x) = 2
    \]

13. Combining the results, we get:
    \[
    c \leq -18 \text{ or } c \geq 2
    \]

The final answer is \(\boxed{c \leq -18 \text{ or } c \geq 2}\).