Только 1 пжлст!!!Упростите и найдете значение алгеброической дроби!!!!

Решите задачу:

$1) \frac{15a^2 - 10ab}{3ab -2b^2} = \frac{5a(3a-2b)}{b(3a-2b)} = \frac{5a}{b} \\ \\ a=-3; b=-0.2 \\ \\ \frac{5*(-3)}{-0.2} = \frac{-15}{-0.2} = \frac{150}{2} = 75 \\ \\ \\ 2)\frac{9c^2-4d^2}{18c^2d -12cd^2} = \frac{(3c)^2-(2d)^2}{6cd(3c -2d)} = \frac{(3c-2d)(3c+2d)}{6cd(3c-2d)} = \frac{3c+2d}{6cd} = \frac{3c}{6cd} + \frac{2d}{6cd} = \\ \\ = \frac{1}{2d} + \frac{1}{3c} \\ \\ c=1 \frac{2}{3} ; d= 0.5 \\ \\ \frac{1}{2*0.5} + \frac{1}{3*1 \frac{2}{3} } = 1 + \frac{1}{5} = 1.2$