ПОМОГИТЕ ДАЮ 40БАЛОВ НА СЕГОДНЯ！

Решите задачу:

$\displaystyle\tt 5) \ \ \frac{x}{12x-3}+\frac{x}{16x-4}=\frac{x}{3(4x-1)}+\frac{x}{4(4x-1)}=\frac{4x+3x}{12(4x-1)}= \\\\\\ {} \ \ \ \ \ =\frac{7x}{48x-12}\\\\\\ 6) \ \ \frac{y}{4x^2-3xy}-\frac{x}{4xy-3y^2}=\frac{y}{x(4x-3y)}-\frac{x}{y(4x-3y)}=\frac{y^2-x^2}{xy(4x-3y)}=\\\\\\{} \ \ \ \ \ =\frac{y^2-x^2}{4x^2y-3xy^2} \\\\\\ 7) \ \ \frac{5}{x-3}+\frac{1}{x+3}+\frac{6}{x^2-9}=\frac{5(x+3)}{x-3}+\frac{1(x-3)}{x+3}+\frac{6}{(x-3)(x+3)}=$

$\displaystyle\tt {} \ \ \ \ \ =\frac{5x+15+x-3+6}{(x-3)(x+3)}=\frac{6x+18}{(x-3)(x+3)}=\frac{6(x+3)}{(x-3)(x+3)}=\frac{6}{x-3}$

ПОМОГИТЕ ДАЮ 40БАЛОВ ** СЕГОДНЯ！