1. Problem: Simplify the expression $$a \cdot \left(\frac{a}{a-2}+1\right) : \left(a + \frac{a^{2}}{2-a}\right)$$. 2. First, rewrite the division as multiplication by the reciprocal: $$a \cdot \left(\frac{a}{a-2}+1\right) \times \frac{1}{a + \frac{a^{2}}{2-a}}$$ 3. Simplify inside the parentheses: $$\frac{a}{a-2} + 1 = \frac{a}{a-2} + \frac{a-2}{a-2} = \frac{a + (a-2)}{a-2} = \frac{2a - 2}{a-2} = \frac{2(a-1)}{a-2}$$ 4. Simplify the denominator expression: Note that $$2 - a = -(a - 2)$$, so $$a + \frac{a^{2}}{2 - a} = a + \frac{a^{2}}{-(a-2)} = a - \frac{a^{2}}{a-2} = \frac{a(a-2)}{a-2} - \frac{a^{2}}{a-2} = \frac{a(a-2) - a^{2}}{a-2} = \frac{a^{2} - 2a - a^{2}}{a-2} = \frac{-2a}{a-2}$$ 5. Now the expression is: $$a \times \frac{2(a-1)}{a-2} \times \frac{1}{\frac{-2a}{a-2}} = a \times \frac{2(a-1)}{a-2} \times \frac{a-2}{-2a}$$ 6. Cancel common factors: $$= a \times \frac{2(a-1)}{\cancel{a-2}} \times \frac{\cancel{a-2}}{-2a} = a \times \frac{2(a-1)}{1} \times \frac{1}{-2a}$$ 7. Cancel $a$ and 2: $$= \cancel{a} \times \frac{2(a-1)}{1} \times \frac{1}{-2 \cancel{a}} = \frac{2(a-1)}{-2} = \frac{\cancel{2}(a-1)}{-\cancel{2}} = -(a-1)$$ 8. Simplify the negative sign: $$-(a-1) = -a + 1 = 1 - a$$ Final answer: $1 - a$