(a-(4a-9/a-2))/(2a-(2a/a-2))

Решите задачу:

$\frac{a-(4a- \frac{9}{a}-2)}{2a-\frac{2a}{a-2}} \\ \\ \\ a-(4a- \frac{9}{a}-2)=a-4a+ \frac{9}{a}+2=-3a+ \frac{9}{a}+2=\frac{-3a^{2}+2a+9}{a} \\ \\ \\ 2a-\frac{2a}{a-2}=\frac{2a(a-2)-2a}{a-2}=\frac{2a^{2}-4a-2a}{a-2}=\frac{2a^{2}-6a}{a-2} \\ \\ \\ \frac{-3a^{2}+2a+9}{a}:\frac{2a^{2}-6a}{a-2}=\frac{-3a^{2}+2a+9}{a}*\frac{a-2}{2a^{2}-6a}=\frac{(-3a^{2}+2a+9)(a-2)}{a(2a^{2}-6a)}= \\ \\ =\frac{-3a^{3}+6a^{2}+2a^{2}-4a+9a-18}{2a^{3}-6a^{2}}=\frac{-3a^{3}+8a^{2}+5a-18}{2a^{3}-6a^{2}}$