Question

Помогите с алгеброй, умоляю, даю 30 баллов

Answer 1

$\dfrac{5}{x^2+5x} +\dfrac{x+15}{25-x^2} = \dfrac{5}{(x+5)}+\dfrac{x+15}{(5-x)(5+x)} = \dfrac{5(5-x)+x(x+15)}{x(5-x)(x+5)} =\\\\= \dfrac{25+10x+x^2}{x(5-x)(5+x)} = \dfrac{(5+x)^2}{x(5-x)(x+5)} = \dfrac{5+x}{x(5-x)} = \dfrac{5+x}{5x-x^2}$

$\dfrac{1}{x+3} + \dfrac{9x}{x^3+27} = \dfrac{1}{x+3} + \dfrac{9x}{(x+3)(x^2-3x+9)} = \dfrac{x^2-3x+9+9x}{(x+3)(x^2-3x+9)} = \\\\=\dfrac{2x^2+6x+9}{(x+3)(x^2-3x+9)} = \dfrac{(x+3)^2}{(x+3)(x^2-3x+9)} = \dfrac{x+3}{x^2-3x+9}$

$\dfrac{2}{x}+\dfrac{12}{x^2-6x} - \dfrac{1-x}{x-6} = \dfrac{2}{x}+\dfrac{6}{x(x-6)}- \dfrac{1-x}{x-6} = \dfrac{2(x-6)+12-x(1-x)}{x(x-6)} = \\\\=\dfrac{2x-12+12-x+x^2}{x(x-6)} = \dfrac{x+x^2}{x(x-6)} = \dfrac{x(1+x)}{x(x-6)} = \dfrac{1+x}{x-6}$