-4/1.2+ -4/2.3+-4/3.4+...+ -4/97.98+-4/98.99

To simplify the expression given, we first notice that we have two identical series:

\[
-4 \left( \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{97.98} + \frac{1}{98.99} \right)
\]

We can denote the series inside the parentheses as \( S \):

\[
S = \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{97.98} + \frac{1}{98.99}
\]

The final expression then becomes:

\[
-4S - 4S = -8S
\]

Now, we must calculate \( S \).

### Analyzing Terms of S

The general term of \( S \) appears to be of the form:

\[
\frac{1}{n(n+1)} \quad \text{where } n \text{ runs from 1 to 98.}
\]

### Calculating S

We can express \( S \) as follows:

\[
S = \sum_{n=1}^{98} \frac{1}{n(n+1)}
\]

Using the technique of partial fractions, we have:

\[
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]

Thus, we can rewrite \( S \):

\[
S = \sum_{n=1}^{98} \left( \frac{1}{n} - \frac{1}{n+1} \right)
\]

### Telescoping Series

This forms a telescoping series:

\[
S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{98} - \frac{1}{99} \right)
\]

When we sum this, all intermediate terms will cancel out, leaving us with:

\[
S = 1 - \frac{1}{99}
\]

Calculating this gives:

\[
S = \frac{99}{99} - \frac{1}{99} = \frac{98}{99}
\]

### Final Calculation

Now substituting \( S \) back into our expression for the total:

\[
-8S = -8 \left( \frac{98}{99} \right) = -\frac{784}{99}
\]

Thus, the final result of the entire expression is:

\[
\boxed{-\frac{784}{99}}
\]