I can now see the image you uploaded. I'll solve the problems step by step.

Problem 1:

$\frac{5}{a^2 + 3a + 2} + \frac{5a + 1}{a^2 + 3a + 2}$

First, notice that both denominators are the same, so we can directly combine the numerators: $\frac{5 + (5a + 1)}{a^2 + 3a + 2} = \frac{5a + 6}{a^2 + 3a + 2}$

Now, factor the denominator: $a^2 + 3a + 2 = (a + 1)(a + 2)$

Thus, the final simplified answer is: $\frac{5a + 6}{(a + 1)(a + 2)}$

Problem 2:

$\frac{6}{x^2 + 8x + 4} - \frac{x + 5}{x^2 + 8x + 4}$

As the denominators are the same, we can subtract the numerators directly: $\frac{6 - (x + 5)}{x^2 + 8x + 4} = \frac{6 - x - 5}{x^2 + 8x + 4} = \frac{1 - x}{x^2 + 8x + 4}$

Thus, the simplified expression is: $\frac{1 - x}{x^2 + 8x + 4}$

Problem 3:

$\frac{10}{x - 3} + \frac{2x + 9}{x - 3} - \frac{5x - 7}{x - 3}$

Since all the denominators are $x - 3$, we can combine the numerators: $\frac{10 + (2x + 9) - (5x - 7)}{x - 3}$ Simplifying the numerator: $10 + 2x + 9 - 5x + 7 = 26 - 3x$

Thus, the final answer is: $\frac{26 - 3x}{x - 3}$

Problem 4:

$\frac{x - 7}{4x^2 - 4} + \frac{8}{4x^2 - 4}$

Factor the denominator: $4x^2 - 4 = 4(x^2 - 1) = 4(x - 1)(x + 1)$

Now, combine the numerators since the denominators are the same: $\frac{(x - 7) + 8}{4(x - 1)(x + 1)} = \frac{x + 1}{4(x - 1)(x + 1)}$

Cancel out the common term $x + 1$: $\frac{1}{4(x - 1)}$

Thus, the final simplified answer is: $\frac{1}{4(x - 1)}$

Would you like further details on any of these steps?

Here are 5 related questions for practice:

1. How do you factor quadratic expressions like $a^2 + 3a + 2$?
2. How do you handle operations involving rational expressions with different denominators?
3. What is the process to combine terms with like denominators in rational expressions?
4. How do you recognize and cancel common factors in fractions?
5. How can you simplify expressions like $\frac{26 - 3x}{x - 3}$?

Tip: Always check if denominators can be factored before adding or subtracting rational expressions. This makes it easier to combine and simplify terms.