Let's simplify the given expression step by step:

$\frac{5x}{x+1} - \frac{x-4}{x} + \frac{5}{x+1}$

Step 1: Combine terms with the same denominator

The denominators in the expression are $x+1$ and $x$. The least common denominator (LCD) is:

$\text{LCD} = x(x+1)$

Rewrite each term with the LCD:

1. $\frac{5x}{x+1} = \frac{5x \cdot x}{x(x+1)} = \frac{5x^2}{x(x+1)}$
2. $\frac{x-4}{x} = \frac{(x-4)(x+1)}{x(x+1)} = \frac{x^2 - 3x - 4}{x(x+1)}$
3. $\frac{5}{x+1} = \frac{5 \cdot x}{x(x+1)} = \frac{5x}{x(x+1)}$

Step 2: Combine all terms under the common denominator

$\frac{5x^2}{x(x+1)} - \frac{x^2 - 3x - 4}{x(x+1)} + \frac{5x}{x(x+1)}$

Combine the numerators: $\text{Numerator} = 5x^2 - (x^2 - 3x - 4) + 5x$

Simplify the numerator: $5x^2 - x^2 + 3x + 4 + 5x = 4x^2 + 8x + 4$

Step 3: Write the simplified fraction

The expression becomes: $\frac{4x^2 + 8x + 4}{x(x+1)}$

Step 4: Factorize the numerator if possible

Factor out 4 from the numerator: $\frac{4(x^2 + 2x + 1)}{x(x+1)}$

Recognize $x^2 + 2x + 1$ as a perfect square: $\frac{4(x+1)^2}{x(x+1)}$

Cancel $x+1$ from the numerator and denominator (valid as long as $x \neq -1$): $\frac{4(x+1)}{x}, \quad \text{for } x \neq -1$

Final Simplified Expression:

$\frac{4(x+1)}{x}$

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Would you like me to elaborate on any of these steps or explore further applications? 😊

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Relative Questions:

1. What is the significance of finding the LCD in rational expressions?
2. How do we determine if terms in the numerator can be factored further?
3. Why must we exclude $x = -1$ when simplifying this expression?
4. Can this type of simplification be applied to integrals or derivatives?
5. How do we handle cases where the LCD becomes more complex?

Tip:

Always check for restrictions on the variable (e.g., where denominators become zero) to ensure your simplified expression is valid.