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Let’s go through the questions step-by-step:

### 1. Simplify each expression and identify values that make the denominator zero.

#### (a) \(\frac{x^2}{x^2}\)
Since anything divided by itself equals 1 (as long as it's not zero),
\[
\frac{x^2}{x^2} = 1 \quad \text{(for } x \neq 0 \text{)}
\]
*Denominator zero value:* \( x = 0 \).

#### (b) \(\frac{x}{x} + \frac{x}{x} + \frac{x}{3}\)
This can be simplified by recognizing that \(\frac{x}{x} = 1\) (for \(x \neq 0\)):
\[
1 + 1 + \frac{x}{3} = 2 + \frac{x}{3}
\]
*Denominator zero value:* \( x = 0 \).

#### (c) \(\frac{x - 2}{x - 2} + \frac{x + 5}{x - 1}\)
Simplifying \(\frac{x - 2}{x - 2} = 1\) (for \(x \neq 2\)):
\[
1 + \frac{x + 5}{x - 1}
\]
*Denominator zero values:* \( x = 2 \) and \( x = 1 \).

#### (d) \(\frac{9 \cdot x}{x \cdot 9}\)
Simplify as follows:
\[
\frac{9x}{9x} = 1 \quad \text{(for } x \neq 0 \text{)}
\]
*Denominator zero value:* \( x = 0 \).

#### (e) \(\frac{h \cdot k + h \cdot k}{h}\)
Factor out \( h \cdot k \) in the numerator:
\[
\frac{2hk}{h} = 2k \quad \text{(for } h \neq 0 \text{)}
\]
*Denominator zero value:* \( h = 0 \).

#### (f) \(\frac{(2m - 5)(m + 6)}{(m + 6)(3m + 1)}\)
Cancel out \((m + 6)\) from the numerator and denominator (for \(m \neq -6\)):
\[
\frac{2m - 5}{3m + 1}
\]
*Denominator zero values:* \( m = -6 \) and \( m = -\frac{1}{3} \).

#### (g) \(\frac{6(n - 2)^2}{3(n - 2)}\)
Cancel \((n - 2)\) from the numerator and denominator (for \(n \neq 2\)):
\[
\frac{6(n - 2)}{3} = 2(n - 2) = 2n - 4
\]
*Denominator zero value:* \( n = 2 \).

#### (h) \(\frac{3 - 2x}{(4x - 1)(3 - 2x)}\)
Cancel \((3 - 2x)\) from the numerator and denominator (for \(x \neq \frac{3}{2}\)):
\[
\frac{1}{4x - 1}
\]
*Denominator zero values:* \( x = \frac{3}{2} \) and \( x = \frac{1}{4} \).

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### 2. Wilson's Question about Rational Expressions

#### (a) Is Wilson correct?
Wilson’s reasoning is incorrect. Simplifying \(\frac{4x + x}{x} = 5\) assumes that \(x \neq 0\). To confirm, we’d graph \(y = \frac{4x + x}{x}\) and \(y = 5\). However, the expression \(\frac{4x + x}{x} = \frac{5x}{x}\) simplifies to \(y = 5\), so the graphs would overlap, showing the statement is true except for \(x = 0\).

#### (b) Comparison of Rational Expressions
A rational expression like \(\frac{x}{x} = 1\) is only valid if \(x \neq 0\). This form is valid when the numerator and denominator are identical and non-zero.

#### (c) Simplification Correctness
The given expression is \(\frac{x^2 + x + 3}{x + 3}\). We check each form:
- \(\frac{x^2 + x + 3}{x + 3} \neq x^2\)
- However, factoring shows \(\frac{(x + 2)(x + 3)}{x + 3} = x + 2\), which is the correct simplification, for \(x \neq -3\).

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Would you like further details on any of these steps, or do you have specific questions?

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### Related Questions
1. What are the steps for simplifying more complex rational expressions?
2. How can factoring help simplify rational expressions with polynomials?
3. When is it appropriate to cancel terms in a rational expression?
4. How do we determine restrictions on the variable in rational expressions?
5. How can graphing help verify the simplification of rational expressions?

### Tip
Always check for values that make the denominator zero before simplifying any rational expression.

Simplify rational expressions and determine values that make the denominator zero. Verify if a rational expression can be simplified to a constant. Check simplification correctness for given expressions.

Learn how to simplify rational expressions, identify values that make the denominator zero, and verify simplification correctness. Ideal for high school students studying algebraic fractions and rational expressions.

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