Упростите. Помогите пожалуйста) Решите мне это.

Question 1

Упростите.
Помогите пожалуйста) Решите мне это.

Question 2

A) $(\frac{1}{a^2-9}:\frac{b-a}{3a^2+9a}-\frac{3a}{9-3b-3a+ab}):\frac{3a}{b^3-27}= \\\\ =(\frac{1}{(a-3)(a+3)}*\frac{3a(a+3)}{b-a}-\frac{3a}{(a-3)(b-3)})*\frac{(b-3)(b^2+3b+9)}{3a}= \\\\ =\frac{3a}{a-3}(\frac{1}{b-a}-\frac{1}{b-3})*\frac{(b-3)(b^2+3b+9)}{3a}= \\\\ =\frac{1}{a-3}*\frac{b-3-(b-a)}{(b-a)(b-3)}*(b-3)(b^2+3b+9)= \\\\ =\frac{b^2+3b+9}{b-a}$

б) $(1+2*\frac{10y^2-3xy}{x^2-3xy})*(\frac{1}{x-5y}+\frac{2y}{(x-5y)^2})= \\\\ =\frac{x^2-3xy+20y^2-6xy}{x(x-3y)}*\frac{x-5y+2y}{(x-5y)^2}= \\\\ =\frac{x^2-9xy+20y^2}{x(x-3y)}*\frac{x-3y}{(x-5y)^2}= \\\\ =\frac{(x^2-10xy+25y^2)+(xy-5y^2)}{x(x-5y)^2}= \\\\ =\frac{(x-5y)^2+y(x-5y)}{x(x-5y)^2}= \\\\ =\frac{(x-5y)(x-4y)}{x(x-5y)^2}= \\\\ =\frac{x-4y}{x(x-5y)}$

в) $\frac{1}{m^2+2mn+2n^2}-\frac{1}{m^2-2mn+2n^2}-\frac{m}{n(m^2+2n^2)}+\frac{m}{n(m^2-2n^2)}= \\\\ =\frac{m^2-2mn+2n^2-(m^2+2mn+2n^2)}{((m^2+2n^2)+2mn)*((m^2+2n^2)-2mn)}-\frac{m}{n}*(\frac{1}{m^2+2n^2}-\frac{1}{m^2-2n^2})= \\\\ =-\frac{4mn}{(m^2+2n^2)^2-4m^2n^2}-\frac{m}{n}*\frac{m^2-2n^2-(m^2+2n^2)}{m^4-4n^4}= \\\\ =-\frac{4mn}{m^4+4m^2n^2+4n^4-4m^2n^2}+\frac{m}{n}*\frac{4n^2}{m^4-4n^4}= \\\\ =-\frac{4mn}{m^4+4n^4}+\frac{4mn}{m^4-4n^4}= \\\\ =4mn*(\frac{1}{m^4-4n^4}-\frac{1}{m^4+4n^4})=$
$=4mn*\frac{m^4+4n^4-(m^4-4n^4)}{(m^4-4n^4)(m^4+4n^4)}= \\\\ =\frac{32mn^5}{m^8-16n^8}$