Срочно памагите надо

Решите задачу:

$\frac{2}{x-3}+ \frac{1}{x+3}+ \frac{2x}{9-x^2}= \frac{2}{x-3}+ \frac{1}{x+3}- \frac{2x}{x^2-9}= \frac{2(x+3)+x-3-2x}{x^2-9}= \frac{x+3}{x^2-9}=\\\\ = \frac{x+3}{(x-3)(x+3)}= \frac{1}{x-3}$

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

$\frac{4x^2+4}{16x^2-9}- \frac{x^2+1}{4x^2-3x}= \frac{4(x^2+1)}{(4x-3)(4x+3)}- \frac{x^2+1}{x(4x-3)}= \\\\ = \frac{4x(x^2+1)-(x^2+1)(4x+3)}{x(4x-3)(4x+3)}= \frac{(x^2+1)(4x-4x-3)}{x(4x-3)(4x+3)}= \frac{-3(x^2+1)}{x(16x^2-9)}= \frac{3x^2+3}{9x-16x^3}$