To solve the equation

\[
\frac{3}{35} + \frac{3}{3} + \frac{3}{99} + \ldots + \frac{3}{x(x+2)} = \frac{27}{95},
\]

let's first analyze the series on the left-hand side.

We can express the general term in the series as:

\[
\frac{3}{n(n + 2)}
\]

where \( n \) takes values starting from 1 and increasing by 2 for each term, leading us to consider terms like \( n = 3, 5, 7, \ldots \) up to \( x \).

Now our goal is to express \( \frac{3}{n(n+2)} \) using partial fractions:

\[
\frac{3}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}.
\]

Multiplying both sides by \( n(n+2) \):

\[
3 = A(n + 2) + Bn.
\]

Setting up the equation gives us:

\[
3 = (A + B)n + 2A.
\]

From this, we can equate coefficients:

1. \( A + B = 0 \)
2. \( 2A = 3 \)

From the second equation, we find:

\[
A = \frac{3}{2}, \quad \text{implying} \quad B = -\frac{3}{2}.
\]

Thus, we have:

\[
\frac{3}{n(n + 2)} = \frac{3/2}{n} - \frac{3/2}{n + 2}.
\]

We can then rewrite our sum:

\[
\sum_{n} \left( \frac{3/2}{n} - \frac{3/2}{n + 2} \right).
\]

This is a telescoping series. The terms will cancel out, leaving us with only the first few terms.

Suppose the last term we are summing is from \( n = x \):

The series would be:

\[
\frac{3}{2} \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \ldots - \frac{1}{x + 2} \right).
\]

Hence:

\[
= \frac{3}{2} \left( 1 - \frac{1}{x + 2} \right).
\]

Thus, our left-hand side becomes:

\[
\frac{3}{2} \left( 1 - \frac{1}{x+2} \right) = \frac{3}{2} \cdot \frac{x + 1}{x + 2} = \frac{3(x+1)}{2(x+2)}.
\]

Setting this equal to the right-hand side:

\[
\frac{3(x+1)}{2(x+2)} = \frac{27}{95}.
\]

Cross-multiplying gives:

\[
3(x + 1) \cdot 95 = 27 \cdot 2(x + 2).
\]

Which simplifies to:

\[
285x + 285 = 54x + 108.
\]

Now, bringing all terms involving \( x \) to one side:

\[
285x - 54x = 108 - 285
\]

This simplifies to:

\[
231x = -177 \Rightarrow x = -\frac{177}{231} = -\frac{59}{77}.
\]

So, the value for \( x \) is:

\[
x = -\frac{59}{77}.
\]

**Final result:**
\[
\boxed{-\frac{59}{77}}.
\]