8 класс пожалйста дроби

Решите задачу:

$2z^3-6x^5=2(z^3-3x^5) \\\ 4a^2- \frac{1}{9} b^2=(2a- \frac{1}{3} b)(2a +\frac{1}{3} b) \\\ q^2-4q^{-5}=q^2(1-4q^{-7}) \\\ 9b^2-3=3(3b^2-1)$

$\frac{x^2-8x+16}{x^2-16}= \frac{(x-4)^2}{(x-4)(x+4)}= \frac{x-4}{x+4} \\\ \frac{16ab^4}{-56ab^3}=- \frac{2b}{7} \\\ \frac{x^2-9}{x^2-6x+9} = \frac{(x-3)(x+3)}{(x-3)^2} = \frac{x+3}{x-3} \\\ \frac{3x-9y}{(3-x)(x-y)}= \frac{3(x-3y)}{(3-x)(x-y)}$
$\frac{3y}{4y-4} + \frac{2y}{5-5y}= \frac{3y}{4(y-1)} - \frac{2y}{5(y-1)}=\frac{15y}{20(y-1)} - \frac{8y}{20(y-1)}=\frac{7y}{20(y-1)}$
$4a- \frac{8a^2}{2a-3} = \frac{8a^2-12a-8a^2}{2a-3} =-\frac{12a}{2a-3}$

$\frac{3x+2}{5x}- \frac{5x+3y}{10xy} - \frac{y-1}{2y} = \frac{6xy+4y}{10xy}- \frac{5x+3y}{10xy} - \frac{5xy-5x}{10xy} = \\\ = \frac{6xy+4y-5x-3y-5xy+5x}{10xy}= \frac{xy+y}{10xy}= \frac{x+1}{10x}=$