Приведите к общему знаменателю дроби:
а) $\frac{x}{4x + x^2}$ и $\frac{4}{3x + 12}$;
б) $\frac{13x}{25 - x^2}$ и $\frac{x - 1}{10 + 2x}$;
в) $\frac{x - 3}{4 - x^2}$ и $\frac{5x}{x^2 - 4}$;
г) $\frac{2}{(x - 3)^2}$ и $\frac{1 + x}{x^2 - 9}$.
$\frac{x}{4x + x^2}$ и $\frac{4}{3x + 12}$
$\frac{x}{4x + x^2} = \frac{x}{x(4 + x)} = \frac{1}{x + 4}$
$\frac{4}{3x + 12} = \frac{4}{3(x + 4)}$
$\frac{1}{x + 4} = \frac{3}{3(x + 4)}$
$\frac{3}{3(x + 4)}$ и $\frac{4}{3(x + 4)}$
$\frac{13x}{25 - x^2}$ и $\frac{x - 1}{10 + 2x}$
$\frac{13x}{25 - x^2} = \frac{13x}{(5 - x)(5 + x)}$
$\frac{x - 1}{10 + 2x} = \frac{x - 1}{2(5 + x)}$
$\frac{13x}{(5 - x)(5 + x)} = \frac{26x}{2(5 - x)(5 + x)}$
$\frac{x - 1}{2(5 + x)} = \frac{(x - 1)(5 - x)}{2(5 + x)(5 - x)}$
$\frac{26x}{2(5 - x)(5 + x)}$ и $\frac{(x - 1)(5 - x)}{2(5 + x)(5 - x)}$
$\frac{x - 3}{4 - x^2}$ и $\frac{5x}{x^2 - 4}$
$\frac{5x}{x^2 - 4} = -\frac{5x}{4 - x^2}$
$\frac{x - 3}{4 - x^2}$ и $-\frac{5x}{4 - x^2}$
$\frac{2}{(x - 3)^2}$ и $\frac{1 + x}{x^2 - 9}$
$\frac{1 + x}{x^2 - 9} = \frac{1 + x}{(x - 3)(x + 3)}$
$\frac{2}{(x - 3)^2} = \frac{2(x + 3)}{(x - 3)^2(x + 3)}$
$\frac{1 + x}{(x - 3)(x + 3)} = \frac{(1 + x)(x - 3)}{(x - 3)^2(x + 3)}$
$\frac{2(x + 3)}{(x - 3)^2(x + 3)}$ и $\frac{(1 + x)(x - 3)}{(x - 3)^2(x + 3)}$
Пожауйста, оцените решение
