To solve the sum of the given fractions, let's break down each component and simplify it appropriately. The fractions given in the problem are:
[tex]\[
\frac{2}{x^2}, \quad \frac{4}{x^2}, \quad \frac{8}{x^2}, \quad \frac{6}{x^2}, \quad \frac{6}{2x^2}, \quad \frac{6}{x^4}
\][/tex]
First, notice that:
- The fractions [tex]\(\frac{2}{x^2}, \frac{4}{x^2}, \frac{8}{x^2}, \frac{6}{x^2}\)[/tex] all have a common denominator [tex]\(x^2\)[/tex].
- The fraction [tex]\(\frac{6}{2x^2}\)[/tex] can be simplified.
For the fraction [tex]\(\frac{6}{2x^2}\)[/tex]:
[tex]\[
\frac{6}{2x^2} = \frac{6 \div 2}{x^2} = \frac{3}{x^2}
\][/tex]
Next, we can write down all the simplified fractions:
[tex]\[
\frac{2}{x^2}, \quad \frac{4}{x^2}, \quad \frac{8}{x^2}, \quad \frac{6}{x^2}, \quad \frac{3}{x^2}, \quad \frac{6}{x^4}
\][/tex]
Notice that [tex]\(\frac{6}{x^4}\)[/tex] is in terms of [tex]\(x^4\)[/tex] instead of [tex]\(x^2\)[/tex], which may differ in influence depending on the context, though it is equivalent if [tex]\(x=1\)[/tex].
Let's add the fractions that have a common denominator ([tex]\(x^2\)[/tex]):
[tex]\[
\frac{2}{x^2} + \frac{4}{x^2} + \frac{8}{x^2} + \frac{6}{x^2} + \frac{3}{x^2} = \frac{2 + 4 + 8 + 6 + 3}{x^2} = \frac{23}{x^2}
\][/tex]
The fraction [tex]\(\frac{6}{x^4}\)[/tex] does not initially fit this sum unless we reconsider each specific value but since all others integrate, it can stay as [tex]\(\frac{6}{x^4}\)[/tex].
So, the major solution components are:
[tex]\[
\left(\frac{2}{x^2}, \frac{4}{x^2}, \frac{8}{x^2}, \frac{6}{x^2}, \frac{3}{x^2}\right)
=
(2.0, 4.0, 8.0, 6.0, 3.0)
\][/tex]
Adding these together yields:
[tex]\[
2.0 + 4.0 + 8.0 + 6.0 + 3.0 = 23.0
\][/tex]
So, the final result for the sum of these fractions is:
[tex]\[
23.0
\][/tex] This sum numerically confirms the calculation and integrates it all for simplified results.
If your initial problem insists on context where [tex]\(x = 1\)[/tex]:
Given
[tex]\[
\sum_{i=1}^{n} \frac{a_{i}}{b_{i}} = \frac{\sum_{i=1}^{n} a_{i}}{\sum_{i=1}^{n} b_{i}}
\][/tex]
Other handled condition retains precise weight as previously discussed ensuring adjusted numeric resultant.
