Let's break down the expression using properties of logarithms and solve it step by step:

The given expression is:

$5\log_{25}(5) - 4\log_{25}(5) + \log_{25}\left(\frac{1}{625}\right)$

Step 1: Combine like terms

The first two terms both involve $\log_{25}(5)$. So, we can simplify them as follows:

$(5 - 4)\log_{25}(5) = \log_{25}(5)$

Step 2: Evaluate $\log_{25}(5)$

Recall that $25 = 5^2$, so:

$\log_{25}(5) = \frac{1}{2}$

Thus, the expression becomes:

$\frac{1}{2} + \log_{25}\left(\frac{1}{625}\right)$

Step 3: Simplify $\log_{25}\left(\frac{1}{625}\right)$

First, express $625$ as a power of $25$. Since $625 = 25^2$, we can rewrite the logarithm as:

$\log_{25}\left(\frac{1}{625}\right) = \log_{25}(25^{-2})$

By the logarithmic power rule, $\log_b(x^n) = n \log_b(x)$, this becomes:

$\log_{25}(25^{-2}) = -2 \log_{25}(25)$

Since $\log_{25}(25) = 1$, the expression simplifies to:

$-2$

Step 4: Combine everything

Now, substitute back into the expression:

$\frac{1}{2} - 2 = \frac{1}{2} - \frac{4}{2} = -\frac{3}{2}$

Thus, the value of the expression is:

$\boxed{-\frac{3}{2}}$

Would you like further details on any of the steps?

Here are 5 related questions for further exploration:

1. How do the logarithmic properties apply when changing bases?
2. Can you explain why $\log_{25}(25)$ equals 1?
3. What is the logarithmic rule for division and powers?
4. How would you solve if the base was different, like $\log_5$?
5. How does this simplify when logarithmic arguments are fractions?

Tip: Always try breaking down the expression using basic logarithmic identities like the product, quotient, and power rules before simplifying further.