Преоброзуйте в дробь 8 класс,помогите срочно надо

А)
$\frac{a+4}{4a} - \frac{a-2}{a^2} = \frac{a(a+4)}{4a*a} - \frac{4(a-2)}{a^2*4} = \frac{a(a+4)- 4(a-2)}{4a^2} = \\ \frac{a^2+4a- 4a+8}{4a^2} = \frac{a^2+8}{4a^2}$

б)
$\frac{3x}{x+3}+ \frac{3x}{x-3}= \frac{3x(x-3)}{(x+3)(x-3)}+ \frac{3x(x+3)}{(x-3)(x+3)}= \\ \frac{3x(x-3)+ 3x(x+3)}{(x-3)(x+3)}= \frac{3x((x-3)+(x+3))}{(x-3)(x+3)}= \frac{3x(x-3+x+3)}{(x-3)(x+3)}= \\ \frac{3x2x}{(x-3)(x+3)}=\frac{6x^2}{x^2-9}$

в)
$\frac{9x^2}{3xy-y^2}- \frac{y}{3x-y} = \frac{9x^2}{y(3x-y)}- \frac{y}{3x-y} = \\ \frac{9x^2}{y(3x-y)}- \frac{y*y}{y(3x-y)} =\frac{9x^2-y^2}{y(3x-y)} =\frac{(3x)^2-y^2}{y(3x-y)} = \\ \frac{(3x-y)(3x+y)}{y(3x-y)} = \frac{3x+y}{y}$

г)
$a-3b+ \frac{6ab}{a-3b}= a* \frac{a-3b}{a-3b}-3b*\frac{a-3b}{a-3b}+ \frac{6ab}{a-3b}= \\ \frac{a(a-3b)-3b(a-3b)+6ab}{a-3b} =\frac{a^2-3ab-3ab+9b^2+6ab}{a-3b} =\frac{a^2+9b^2}{a-3b}$