Question

Желательно побыстрее, спасибо

Answer 1

$\left( \dfrac{a+x}{x} - \dfrac{2x}{x-a} \right) \cdot \dfrac{x-a}{a^2+x^2} = \dfrac{(x+a)(x-a) - 2x^2}{x(x-a)} \cdot \dfrac{x-a}{a^2+x^2} = \dfrac{x^2-a^2-2x^2}{x(x-a)} \cdot \dfrac{x-a}{a^2+x^2} = -\dfrac{a^2+x^2}{x(x-a)} \cdot \dfrac{x-a}{a^2+x^2} = -\dfrac{1}{x}$

$\left( \dfrac{x-2}{3x+6} + \dfrac{1}{x^2-4} + \dfrac{x-6}{6-3x} \right) \cdot \dfrac{9x^2-36}{32} = \left( \dfrac{(x-2)(3x-6) - (x-6)(3x+6)}{(3x+6)(3x-6)} + \dfrac{1}{x^2-4} \right) \cdot \dfrac{9(x^2-4)}{32} = \left( \dfrac{3x^2-12x+12 - 3x^2+12x+36}{9x^2-36} + \dfrac{1}{x^2-4}\right) \cdot \dfrac{9(x^2-4)}{32} = \left( \dfrac{48}{9(x^2-4)} + \dfrac{1}{x^2-4} \right)\cdot \dfrac{9(x^2-4)}{32} = \dfrac{48+9}{9(x^2-4)}\cdot \dfrac{9(x^2-4)}{32} = \dfrac{57}{32}$