The expression you provided is:

\[
\frac{a + 2}{a + 1} + \frac{7}{a^2 - 5a - 6}
\]

Let's simplify it step by step.

### Step 1: Factorize the denominator in the second term
We can factor \(a^2 - 5a - 6\).

To factor \(a^2 - 5a - 6\), we look for two numbers that multiply to \(-6\) and add up to \(-5\). These numbers are \(-6\) and \(1\), so we can factor the quadratic as:

\[
a^2 - 5a - 6 = (a - 6)(a + 1)
\]

Now, the expression becomes:

\[
\frac{a + 2}{a + 1} + \frac{7}{(a - 6)(a + 1)}
\]

### Step 2: Find the common denominator
The common denominator between \(\frac{a + 2}{a + 1}\) and \(\frac{7}{(a - 6)(a + 1)}\) is \((a + 1)(a - 6)\).

Rewrite the first term with the common denominator:

\[
\frac{a + 2}{a + 1} = \frac{(a + 2)(a - 6)}{(a + 1)(a - 6)}
\]

Now the expression becomes:

\[
\frac{(a + 2)(a - 6)}{(a + 1)(a - 6)} + \frac{7}{(a + 1)(a - 6)}
\]

### Step 3: Combine the terms
Since the denominators are the same, we can combine the numerators:

\[
\frac{(a + 2)(a - 6) + 7}{(a + 1)(a - 6)}
\]

### Step 4: Expand the numerator
Expand \((a + 2)(a - 6)\):

\[
(a + 2)(a - 6) = a^2 - 6a + 2a - 12 = a^2 - 4a - 12
\]

Now, the expression becomes:

\[
\frac{a^2 - 4a - 12 + 7}{(a + 1)(a - 6)}
\]

Simplify the numerator:

\[
a^2 - 4a - 12 + 7 = a^2 - 4a - 5
\]

### Final expression:
The simplified form of the expression is:

\[
\frac{a^2 - 4a - 5}{(a + 1)(a - 6)}
\]

This is the simplified result.

Would you like further details or clarifications on any part of the solution?

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Here are five related questions:
1. How do you factor quadratic expressions like \(a^2 - 5a - 6\)?
2. What are common denominators, and how are they used in algebraic fractions?
3. How do you simplify complex rational expressions?
4. How do you expand binomials like \((a + 2)(a - 6)\)?
5. Can this expression be simplified further if \(a\) has a specific value?

**Tip:** Always look for factorization opportunities when simplifying algebraic fractions.

Learn how to simplify the rational expression (a + 2)/(a + 1) + 7/(a^2 - 5a - 6) by factoring the quadratic denominator and combining terms using a common denominator. This solution is suitable for students in Grades 9-12.

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