Помогите у меня получается не равное тождеству ((

Question 1

Question 2

$(\frac{a-b}{a^2+ab}-\frac{a}{ab+b^2}):(\frac{1}{a+b}-\frac{b^2}{ab^2-a^3})=\\\ \\\ =(\frac{a-b}{a(a+b)}-\frac{a}{b(a+b)}):(\frac{1}{a+b}-\frac{b^2}{a(b^2-a^2)})=\frac{b(a-b)-a^2}{ab(a+b)}:\frac{a(b-a)-b^2}{a(b-a)(b+a)}=\\\ \\\ =\frac{ab-b^2-a^2}{ab(a+b)}*\frac{a(b-a)(b+a)}{ab-a^2-b^2}=\frac{b-a}{b}$

$a^2+9a+\frac{a^2-9}{a^2-2a+4}:\frac{a+3}{9a^3+72}=\\\ \\\ =a^2+9a+\frac{(a-3)(a+3)}{a^2-2a+4}:\frac{a+3}{9(a^3+8)}=\\\ \\\ =a^2+9a+\frac{(a-3)(a+3)}{a^2-2a+4}*\frac{9(a+2)(a^2-2a+4)}{a+3}=\\\ \\\ =a^2+9a+9(a-3)(a+2)=a^2+9a+9(a^2-a-6)=\\\ \\\ =a^2+9a+9a^2-9a-54=10a^2-54$

$(\frac{4a}{a^2-1}+\frac{a-1}{a+1})*\frac{2a}{a+1}-\frac{a}{a-1}-\frac{a}{a^2-1}=\\\ \\\ =(\frac{4a}{(a-1)(a+1)}+\frac{a-1}{a+1})*\frac{2a}{a+1}-\frac{a}{a-1}-\frac{a}{(a-1)(a+1)}=\\\ \\\ =\frac{4a+(a-1)^2}{(a-1)(a+1)}*\frac{2a}{a+1}-\frac{a(a+1)}{(a-1)(a+1)}-\frac{a}{(a-1)(a+1)}=\\\ \\\ =\frac{4a+a^2-2a+1}{(a-1)(a+1)}*\frac{2a}{a+1}-\frac{a^2+a}{(a-1)(a+1)}-\frac{a}{(a-1)(a+1)}=\\\ \\\ =\frac{a^2-2a+1}{(a-1)(a+1)}*\frac{2a}{a+1}-\frac{a^2+a}{(a-1)(a+1)}-\frac{a}{(a-1)(a+1)}=\\\ \\\$

$=\frac{(a+1)^2}{(a-1)(a+1)}*\frac{2a}{a+1}-\frac{a^2+a}{(a-1)(a+1)}-\frac{a}{(a-1)(a+1)}=\\\ \\\ =\frac{2a(a+1)}{(a-1)(a+1)}-\frac{a^2+a}{(a-1)(a+1)}-\frac{a}{(a-1)(a+1)}=\\\ \\\ =\frac{2a^2+2a-a^2-a-a}{(a-1)(a+1)}= \frac{a^2}{(a-1)(a+1)}=\frac{a^2}{a^2-1}$