Пожалуйста помогите мне срочно нужно

$1) ...= \frac{3-2b}{a+b} * \frac{9(b+a)}{(2b-3)(2b+3)}- \frac{6b}{2b+3}= \frac{9(3-2b)-6b(2b-3)}{(2b-3)(2b+3)}=\frac{(9+6b)(3-2b)}{(2b-3)(2b+3)} =$ $\frac{3(3+2b)(3-2b)}{(2b+3)(2b-3)}= \frac{-3(2b-3)}{2b-3}=-3$

$2) ....= \frac{(4a+5c)(4a-5c)*3c}{c(a+c)(5c-4a)} + \frac{13a+14c}{a+c}= \frac{-3(5c-4a)(4a+5c)}{a+c}+ \frac{13a+14c}{a+c} =$ $\frac{-12a-15c+13a+14c}{a+c}= \frac{a-c}{a+c}$

3)....... $= (\frac{b+2}{b^{2}-3b+9}- \frac{6}{(b+3)(b^{2}-3b+9)})* \frac{2(b^{2}-3b+9)}{3(b+5)}= \frac{(b+2)(b+3)-6}{(b+3)(b^{2}-3b+9)}* \frac{2(b^{2}-3b+9) }{3(b+5)}=$ $\frac{2b(b+8)}{3(b+3)(b+5)}$