Let's simplify the given expression:

\[
\frac{a^2b^2}{2ab} - \frac{a^2}{a - b}
\]

### Step 1: Simplify the first term

The first term is \(\frac{a^2b^2}{2ab}\). We can simplify it by canceling out common factors:

\[
\frac{a^2b^2}{2ab} = \frac{a^2 \cdot b^2}{2 \cdot a \cdot b} = \frac{ab}{2}
\]

So, the first term simplifies to \(\frac{ab}{2}\).

### Step 2: Rewrite the expression

Now the expression becomes:

\[
\frac{ab}{2} - \frac{a^2}{a - b}
\]

### Step 3: Find a common denominator

To subtract these two terms, we need a common denominator. The second term has a denominator of \(a - b\), and the first term can be written with the denominator \(2(a - b)\):

\[
\frac{ab}{2} = \frac{ab(a - b)}{2(a - b)}
\]

Now the expression looks like:

\[
\frac{ab(a - b)}{2(a - b)} - \frac{2a^2}{2(a - b)}
\]

### Step 4: Combine the terms

Now that the denominators are the same, we can combine the numerators:

\[
\frac{ab(a - b) - 2a^2}{2(a - b)}
\]

### Step 5: Expand and simplify the numerator

Expand the numerator:

\[
ab(a - b) = a^2b - ab^2
\]

So the numerator becomes:

\[
a^2b - ab^2 - 2a^2
\]

### Final expression

The simplified form of the original expression is:

\[
\frac{a^2b - ab^2 - 2a^2}{2(a - b)}
\]

This is the final simplified form of the given expression.

Would you like further details or explanations?

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Here are five questions related to this concept:
1. How do you find the least common denominator for rational expressions?
2. What are the key steps in simplifying complex fractions?
3. How can factoring help in simplifying rational expressions?
4. How do you handle subtraction of fractions with different denominators?
5. What role do negative signs play when simplifying algebraic expressions?

**Tip:** Always factor and cancel common terms where possible before attempting to find a common denominator in algebraic fractions.

This solution explains how to simplify the rational expression a²b²/2ab - a²/(a-b) step-by-step. It covers algebraic concepts such as finding a common denominator and combining terms. Suitable for students in Grades 9-11.

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