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We are given the equation to prove:

\[
\frac{(x^2 - 1) e^{\ln x + \ln y}}{x^2 y - xy} = x + 1
\]
where \(x > 1\) and \(y > 1\).

Let's break this down step by step.

### Step 1: Simplify the exponential expression

We have the term \( e^{\ln x + \ln y} \). Using the property of logarithms, \(\ln x + \ln y = \ln(xy)\), so:
\[
e^{\ln x + \ln y} = e^{\ln(xy)} = xy
\]
Thus, the equation becomes:
\[
\frac{(x^2 - 1) xy}{x^2 y - xy}
\]

### Step 2: Simplify the denominator

The denominator can be factored as:
\[
x^2 y - xy = xy(x - 1)
\]

Now the equation becomes:
\[
\frac{(x^2 - 1) xy}{xy(x - 1)}
\]

### Step 3: Simplify the numerator

The numerator is \((x^2 - 1) xy\). Using the difference of squares, \(x^2 - 1 = (x - 1)(x + 1)\), so the numerator becomes:
\[
(x - 1)(x + 1) xy
\]

### Step 4: Cancel common terms

Now the equation looks like:
\[
\frac{(x - 1)(x + 1) xy}{xy(x - 1)}
\]

We can cancel the common terms \(xy\) and \((x - 1)\) from the numerator and denominator:
\[
\frac{(x + 1)}{1} = x + 1
\]

Thus, we have proven that:
\[
\frac{(x^2 - 1) e^{\ln x + \ln y}}{x^2 y - xy} = x + 1
\]

### Final Answer:
The proof is complete, and the given equation holds.

---

Do you have any questions or need further clarification?

Here are five related questions to deepen your understanding:
1. How does the property \(\ln a + \ln b = \ln(ab)\) apply in logarithmic manipulations?
2. What is the difference of squares, and how is it used in factoring \(x^2 - 1\)?
3. Why are we allowed to cancel terms like \((x - 1)\) in this proof?
4. Can you apply this method to other algebraic identities involving exponentials and logarithms?
5. How do logarithmic and exponential functions interact in solving equations?

**Tip:** Always check for opportunities to factor expressions and simplify terms before expanding them in equations!

Prove that: \( \frac{(x^2 - 1)e^{\ln x + \ln y}}{x^2 y - xy} = x + 1 \), where \(x > 1, y > 1\).

Learn how to prove the equation \( \frac{(x^2 - 1)e^{\ln x + \ln y}}{x^2 y - xy} = x + 1 \), by simplifying logarithmic and exponential expressions, and applying the difference of squares. Suitable for students in Grades 10-12.

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