Question: Perform the indicated operations. \frac{2x + 5}{x + 5} + \frac{6}{x} - \frac{25}{x^2 + 5x}…

Perform the indicated operations. \frac{2x + 5}{x + 5} + \frac{6}{x} - \frac{25}{x^2 + 5x} \frac{2x + 5}{x + 5} + \frac{6}{x} - \frac{25}{x^2 + 5x} = \boxed{\text{(Simplify your answer.)}}

Solution

To solve the problem, we need to perform the indicated operations on the fractions:

\[ \frac{2x+5}{x+5} + \frac{6}{x} - \frac{25}{x^2 + 5x} \]

First, let’s find a common denominator. The denominators are \(x + 5\), \(x\), and \(x^2 + 5x\). The least common denominator (LCD) is \(x(x+5)\).

Convert each fraction to have the LCD as the denominator:

\[ \frac{2x+5}{x+5} = \frac{2x+5}{x+5} \cdot \frac{x}{x} = \frac{x(2x+5)}{x(x+5)} \]

\[ \frac{6}{x} = \frac{6}{x} \cdot \frac{x+5}{x+5} = \frac{6(x+5)}{x(x+5)} \]

\[ \frac{25}{x^2 + 5x} = \frac{25}{x(x+5)} \]

Now the expression becomes:

\[ \frac{x(2x+5)}{x(x+5)} + \frac{6(x+5)}{x(x+5)} - \frac{25}{x(x+5)} \]

Combine the fractions:

\[ \frac{x(2x+5) + 6(x+5) - 25}{x(x+5)} \]

Simplify the numerator:

\[ x(2x+5) = 2x^2 + 5x \]

\[ 6(x+5) = 6x + 30 \]

Combine and simplify:

\[ 2x^2 + 5x + 6x + 30 - 25 = 2x^2 + 11x + 5 \]

The expression is now:

\[ \frac{2x^2 + 11x + 5}{x(x+5)} \]

This is the simplified form of the expression.