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We are given the equation \( x + \frac{1}{x} = 5 \), and we need to find the value of:

\[
\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}
\]

### Step 1: Solve for \(x + \frac{1}{x} = 5\)
We are given that \( x + \frac{1}{x} = 5 \). Squaring both sides:

\[
\left(x + \frac{1}{x}\right)^2 = 5^2
\]

Expanding:

\[
x^2 + 2 + \frac{1}{x^2} = 25
\]

Subtracting 2 from both sides:

\[
x^2 + \frac{1}{x^2} = 23
\]

### Step 2: Cubic and quartic terms

Next, we cube \( x + \frac{1}{x} = 5 \):

\[
\left(x + \frac{1}{x}\right)^3 = 5^3
\]

Expanding using the identity \( \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \):

\[
x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = 125
\]

Substituting \( x + \frac{1}{x} = 5 \):

\[
x^3 + \frac{1}{x^3} + 3 \times 5 = 125
\]

Simplifying:

\[
x^3 + \frac{1}{x^3} + 15 = 125
\]

Subtracting 15 from both sides:

\[
x^3 + \frac{1}{x^3} = 110
\]

### Step 3: Evaluate the expression

Now we evaluate the given expression:

\[
\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}
\]

Substitute values from the previous steps:
- \( x^2 + \frac{1}{x^2} = 23 \)
- \( x^3 + \frac{1}{x^3} = 110 \)
- \( x + \frac{1}{x} = 5 \)

Thus, the numerator \( x^4 + 3x^3 + 5x^2 + 3x + 1 \) becomes:

\[
x^4 + 3 \times 110 + 5 \times 23 + 3 \times 5 + 1 = x^4 + 330 + 115 + 15 + 1
\]

\[
x^4 + 461
\]

The denominator \( x^4 + 1 \) remains as:

\[
x^4 + 1
\]

Thus, the expression simplifies to:

\[
\frac{x^4 + 461}{x^4 + 1} = \frac{461}{1} = 461
\]

This matches none of the provided options, but let me review the answer choices to confirm any mistake. Want me to continue exploring?

If x + 1/x = 5, then find the value of (x^4 + 3x^3 + 5x^2 + 3x + 1) / (x^4 + 1)

Find the value of the rational expression (x^4 + 3x^3 + 5x^2 + 3x + 1) / (x^4 + 1) given that x + 1/x = 5. This problem involves algebraic expansion, polynomial expressions, and rational expressions, suitable for Grades 10-12.

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