To solve the equation \( \frac{1}{2} + \left( \frac{5}{6} + \frac{11}{12} + \frac{19}{20} + \frac{29}{30} + \frac{41}{42} + \frac{55}{56} + \frac{71}{72} + \frac{89}{90} \right) x = \frac{81}{10} \), we need to follow these steps:

1. **Combine the fractions inside the parentheses:**

First, find a common denominator for the fractions inside the parentheses. The denominators are 6, 12, 20, 30, 42, 56, 72, and 90. The least common multiple (LCM) of these numbers is 2520.

Convert each fraction to have the denominator 2520:
\[
\frac{5}{6} = \frac{5 \times 420}{6 \times 420} = \frac{2100}{2520}
\]
\[
\frac{11}{12} = \frac{11 \times 210}{12 \times 210} = \frac{2310}{2520}
\]
\[
\frac{19}{20} = \frac{19 \times 126}{20 \times 126} = \frac{2394}{2520}
\]
\[
\frac{29}{30} = \frac{29 \times 84}{30 \times 84} = \frac{2436}{2520}
\]
\[
\frac{41}{42} = \frac{41 \times 60}{42 \times 60} = \frac{2460}{2520}
\]
\[
\frac{55}{56} = \frac{55 \times 45}{56 \times 45} = \frac{2475}{2520}
\]
\[
\frac{71}{72} = \frac{71 \times 35}{72 \times 35} = \frac{2485}{2520}
\]
\[
\frac{89}{90} = \frac{89 \times 28}{90 \times 28} = \frac{2492}{2520}
\]

Now, sum these fractions:
\[
\frac{2100 + 2310 + 2394 + 2436 + 2460 + 2475 + 2485 + 2492}{2520} = \frac{19152}{2520}
\]

Simplify the fraction:
\[
\frac{19152}{2520} = \frac{2394}{315} = \frac{798}{105} = \frac{266}{35} \approx 7.6
\]

2. **Substitute back into the equation:**

\[
\frac{1}{2} + \left( \frac{266}{35} \right) x = \frac{81}{10}
\]

3. **Isolate \( x \):**

First, subtract \(\frac{1}{2}\) from both sides:
\[
\left( \frac{266}{35} \right) x = \frac{81}{10} - \frac{1}{2}
\]

Find a common denominator for the right-hand side:
\[
\frac{81}{10} = \frac{81 \times 5}{10 \times 5} = \frac{405}{50}
\]
\[
\frac{1}{2} = \frac{1 \times 25}{2 \times 25} = \frac{25}{50}
\]
\[
\frac{405}{50} - \frac{25}{50} = \frac{380}{50} = \frac{38}{5}
\]

Now, solve for \( x \):
\[
\left( \frac{266}{35} \right) x = \frac{38}{5}
\]
\[
x = \frac{\frac{38}{5}}{\frac{266}{35}} = \frac{38}{5} \times \frac{35}{266} = \frac{38 \times 35}{5 \times 266} = \frac{1330}{1330} = 1
\]

Thus, the solution to the equation is \( x = 1 \).