Question: Add. Simplify, if possible. \frac{1}{v + 4} + \frac{4}{5v} + \frac{4}{v^2 + 4v} The sum is…

Add. Simplify, if possible. \frac{1}{v + 4} + \frac{4}{5v} + \frac{4}{v^2 + 4v}

The sum is \boxed{ } (Simplify your answer.)

Solution

To solve the problem, we need to add and simplify the given rational expressions:

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

First, find a common denominator. The least common denominator (LCD) of these fractions is \(v^2 + 4v\).

Rewrite each fraction with this common denominator.

\[ \frac{1}{v+4} = \frac{1 \cdot v}{(v+4)(v)} = \frac{v}{v^2 + 4v} \]

\[ \frac{4}{5v} = \frac{4 \cdot (v+4)}{5v \cdot (v+4)} = \frac{4v + 16}{5v^2 + 20v} \]

\[ \frac{4}{v^2 + 4v} = \frac{4}{v^2 + 4v} \]

Now, adjust each fraction to have the LCD:

\[ \frac{1}{v+4} = \frac{v}{v^2 + 4v} \]

\[ \frac{4}{5v} = \frac{4(v+4)}{v^2 + 4v} \]

Now add these fractions:

\[ \frac{v}{v^2 + 4v} + \frac{4v + 16}{v^2 + 4v} + \frac{4}{v^2 + 4v} \]

Combine the numerators:

\[ \frac{v + 4v + 16 + 4}{v^2 + 4v} = \frac{5v + 20}{v^2 + 4v} \]

Factor out the greatest common factor from the numerator:

\[ \frac{5(v + 4)}{v^2 + 4v} \]

Since there is no further simplification possible, the simplified sum is:

\[ \frac{5(v + 4)}{v^2 + 4v} \]