Преобразуйте в дробь выражение:
а) $\frac{b - 6}{4 - b^2} + \frac{2}{2b - b^2}$;
б) $\frac{b}{ab - 5a^2} - \frac{15b - 25a}{b^2 - 25^2}$;
в) $\frac{x - 12a}{x^2 - 16a^2} - \frac{4a}{4ax - x^2}$;
г) $\frac{a - 30y}{a^2 - 100y^2} - \frac{10y}{10ay - a^2}$.
$\frac{b - 6}{4 - b^2} + \frac{2}{2b - b^2} = \frac{b - 6}{(2 - b)(2 + b)} + \frac{2}{b(2 - b)} = \frac{b(b - 6) + 2(2 + b)}{b(2 - b)(2 + b)} = \frac{b^2 - 6b + 4 + 2b}{b(2 - b)(2 + b)} = \frac{b^2 - 4b + 4}{b(2 - b)(2 + b)} = \frac{(b - 2)^2}{b(2 - b)(2 + b)} = \frac{(2 - b)^2}{b(2 - b)(2 + b)} = \frac{2 - b}{b(2 + b)}$
$\frac{b}{ab - 5a^2} - \frac{15b - 25a}{b^2 - 25^2} = \frac{b}{a(b - 5a)} - \frac{15b - 25a}{(b - 5a)(b + 5a)} = \frac{b(b + 5a) - a(15b - 25a)}{a(b - 5a)(b + 5a)} = \frac{b^2 + 5ab - 15ab + 25a^2}{a(b - 5a)(b + 5a)} = \frac{b^2 - 10ab + 25a^2}{a(b - 5a)(b + 5a)} = \frac{(b - 5a)^2}{a(b - 5a)(b + 5a)} = \frac{b - 5a}{a(b + 5a)}$
$\frac{x - 12a}{x^2 - 16a^2} - \frac{4a}{4ax - x^2} = \frac{x - 12a}{(x - 4a)(x + 4a)} + \frac{4a}{x(x - 4a)} = \frac{x(x - 12a) + 4a(x + 4a)}{x(x - 4a)(x + 4a)} = \frac{x^2 - 12ax + 4ax + 16a^2}{x(x - 4a)(x + 4a)} = \frac{x^2 - 8ax + 16a^2}{x(x - 4a)(x + 4a)} = \frac{(x - 4a)^2}{x(x - 4a)(x + 4a)} = \frac{x - 4a}{x(x + 4a)}$
$\frac{a - 30y}{a^2 - 100y^2} - \frac{10y}{10ay - a^2} = \frac{a - 30y}{(a - 10y)(a + 10y)} + \frac{10y}{a(a - 10y)} = \frac{a(a - 30y) + 10y(a + 10y)}{a(a - 10y)(a + 10y)} = \frac{a^2 - 30ay + 10ay + 100y^2}{a(a - 10)(a + 10y)} = \frac{a^2 - 20ay + 100y^2}{a(a - 10y)(a + 10y)} = \frac{(a - 10y^2)}{a(a - 10y)(a + 10y)} = \frac{a - 10y}{a(a + 10y)}$
Пожауйста, оцените решение