Можно 4й пример 474 номера?

Решите задачу:

$\frac{18}{9-4a^2} -\frac{4}{2a+3} +\frac{3}{2a-3} =\\\\=\frac{18}{(3-2a)(3+2a)} -\frac{4}{2a+3} +\frac{3}{2a-3} =\\\\=\frac{-18}{(2a-3)(2a+3)} -\frac{4}{2a+3} +\frac{3}{2a-3} =\\\\=\frac{-18}{(2a-3)(2a+3)} -\frac{4(2a-3)}{(2a+3)(2a-3)} +\frac{3(2a+3)}{(2a-3)(2a+3)} =\\\\=\frac{-18}{(2a-3)(2a+3)} -\frac{8a-12}{(2a+3)(2a-3)} +\frac{6a+9}{(2a-3)(2a+3)} =\\\\=\frac{-18-(8a-12)+(6a+9)}{(2a-3)(2a+3)} =\frac{-18-8a+12+6a+9}{(2a-3)(2a+3)} =\\\\=\frac{-2a+3}{(2a-3)(2a+3)}=\frac{-2a+3}{-(-2a+3)(2a+3)}=$

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

$-\frac{1}{2*(-0,6)+3}=-\frac{1}{-1,2+3} =-\frac{1}{1,8} =-\frac{1}{\frac{18}{10} } =-\frac{10}{18} =-\frac{5}{9}$