Алгебра, простой пример

Решите задачу:

$(\frac{4}{a^{1,5}-8}-\frac{a^{0,5}-2}{a+2a^{0,5}+4})*\frac{a^2-8a^{0,5}}{a-16}\\ \\ 1)\frac{4}{a^{1,5}-8}-\frac{a^{0,5}-2}{a+2a^{0,5}+4} =\frac{4}{(a^{0,5}-2)(a+2a^{0,5}+4)}-\frac{a^{0,5}-2}{a+2a^{0,5}+4}= \\ \\ =\frac{4-(a^{0,5}-2)^2}{(a^{0,5}-2)(a+2a^{0,5}+4)} =\frac{4-(a-4a^{0,5}+4)}{(a^{0,5}-2)(a+2a^{0,5}+4)} =\frac{4-a+4a^{0,5}-4}{(a^{0,5}-2)(a+2a^{0,5}+4)}= \\ \\ =\frac{-a+4a^{0,5}}{(a^{0,5}-2)(a+2a^{0,5}+4)}=-\frac{a^{0,5}(a^{0,5}-4)}{a^{1,5}-8}$

$2)-\frac{a^{0,5}(a^{0,5}-4)}{a^{1,5}-8}*\frac{a^2-8a^{0,5}}{a-16} =-\frac{a^{0,5}(a^{0,5}-4)}{a^{1,5}-8}*\frac{a^{0,5}(a^{1,5}-8)}{(a^{0,5}-4)(a^{0,5}+4)}= \\ \\ =-\frac{a^{0,5}}{1}*\frac{a^{0,5}}{a^{0,5}+4}=-\frac{a}{a^{0,5}+4} \\ \\ 3)-\frac{a}{a^{0,5}+4}-\frac{4a^{0,5}}{a^{0,5}+4}=\frac{-a-4a^{0,5}}{a^{0,5}+4} =-\frac{a+4a^{0,5}}{a^{0,5}+4}=-\frac{a^{0,5}(a^{0,5}+4)}{a^{0,5}+4}=-a^{0,5}$