The problem asks to simplify the following expression: $\frac{\frac{a}{a-3}}{1-\frac{a^2-2a}{a-3}} \div \frac{a}{a-3}$

1. Problem Description

The problem asks to simplify the following expression:

$\frac{\frac{a}{a-3}}{1-\frac{a^2-2a}{a-3}} \div \frac{a}{a-3}$

2. Solution Steps

First, simplify the denominator of the main fraction:

$1-\frac{a^2-2a}{a-3} = \frac{a-3}{a-3} - \frac{a^2-2a}{a-3} = \frac{a-3-(a^2-2a)}{a-3} = \frac{a-3-a^2+2a}{a-3} = \frac{-a^2+3a-3}{a-3}$

Now, the main fraction becomes:

$\frac{\frac{a}{a-3}}{\frac{-a^2+3a-3}{a-3}} = \frac{a}{a-3} \cdot \frac{a-3}{-a^2+3a-3} = \frac{a}{-a^2+3a-3}$

Next, we have to divide the result by $\frac{a}{a-3}$. This is equivalent to multiplying by the reciprocal:

$\frac{a}{-a^2+3a-3} \div \frac{a}{a-3} = \frac{a}{-a^2+3a-3} \cdot \frac{a-3}{a} = \frac{a(a-3)}{a(-a^2+3a-3)}$

We can cancel out $a$ from the numerator and denominator:

$\frac{a(a-3)}{a(-a^2+3a-3)} = \frac{a-3}{-a^2+3a-3}$

Therefore the simplified expression is $\frac{a-3}{-a^2+3a-3}$.

Multiplying top and bottom by $-1$ gives $\frac{3-a}{a^2-3a+3}$

3. Final Answer

$\frac{3-a}{a^2-3a+3}$