Допооооможіть!!!!!!! даю 100 балов

Решите задачу:

$\displaystyle\tt\boxed{1}\\\\ \frac{63x^2y^2}{42x^6y^5}= \frac{3}{2x^4y^3}\\\\\\ \frac{49-a^2}{3a^2-21a}=\frac{(7-a)(7+a)}{-3a(7-a)}=-\frac{7+a}{3a}\\\\\\ \frac{(5-x)^2}{3x^2-30x+75}=\frac{(5-x)^2}{3(25-10x+x^2)}=\frac{(5-x)^2}{3(5-x)^2}=\frac{1}{3}$

$\displaystyle\tt\boxed{2}\\\\ \frac{d}{c-d}=\frac{d(c+d)}{(c-d)(c+d)}= \frac{cd+d^2}{c^2-d^2}$

$\displaystyle\tt\boxed{3}\\\\ \frac{m^2+6m+9}{9-m^2}= \frac{(m+3)^2}{(3-m)(3+m)}=\frac{m+3}{3-m}=\frac{7+3}{3-7}=\frac{10}{-4}=-2.5$