Представьте в виде дроби:
а) $\frac{4}{y + 2} - \frac{3}{y - 2} + \frac{12}{y^2 - 4}$;
б) $\frac{a}{a - 6} - \frac{3}{a + 6} + \frac{a^2}{36 - a^2}$;
в) $\frac{x^2}{(x - y)^2} - \frac{x + y}{2x - 2y}$;
г) $\frac{b}{(a - b)^2} - \frac{a + b}{b^2 - ab}$.
$\frac{4}{y + 2} - \frac{3}{y - 2} + \frac{12}{y^2 - 4} = \frac{4(y - 2) - 3(y + 2) + 12}{(y + 2)(y - 2)} = \frac{y - 2}{(y + 2)(y - 2)} = \frac{1}{y + 2}$
$\frac{a}{a - 6} - \frac{3}{a + 6} + \frac{a^2}{36 - a^2} = \frac{a(a + 6) - 3(a - 6) - a^2}{(a - 6)(a + 6)} = \frac{a^2 + 6a - 3a + 18 - a^2}{(a - 6)(a + 6)} = \frac{3(a + 6)}{(a - 6)(a+ 6)} = \frac{3}{a - 6}$
$\frac{x^2}{(x - y)^2} - \frac{x + y}{2x - 2y} = \frac{2x^2 - (x + y)(x - y)}{2(x - y)^2} = \frac{x^2 + y^2}{2(x - y)^2}$
$\frac{b}{(a - b)^2} - \frac{a + b}{b^2 - ab} = \frac{b}{(a - b)^2} + \frac{a + b}{b(a - b)} = \frac{b^2 + (a + b)(a - b)}{b(a - b)^2} = \frac{a^2}{b(a - b)^2}$
Пожауйста, оцените решение
