Сократите дробь:
а) $\frac{ax + ay - bx - by}{x^2 + xy}$;
б) $\frac{a^2 - 2ab + b^2}{ac - bc + bd - ad}$;
в) $\frac{ax - ay - x^2 + xy}{ax - a^2}$;
г) $\frac{b^2 - 2b + 1}{c - bc + a - ab}$.
$\frac{ax + ay - bx - by}{x^2 + xy} = \frac{(ax + ay) - (bx + by)}{x(x + y)} = \frac{a(x + y) - b(x + y)}{x(x + y)} = \frac{(a - b)(x + y)}{x(x + y)} = \frac{a - b}{x}$
$\frac{a^2 - 2ab + b^2}{ac - bc + bd - ad} = \frac{(a - b)^2}{(ac - bc) + (bd - ad)} = \frac{(a - b)^2}{c(a - b) + d(b - a)} = \frac{(a - b)^2}{c(a - b) - d(a - b)} = \frac{(a - b)^2}{(a - b)(c - d)} = \frac{a - b}{c - d}$
$\frac{ax - ay - x^2 + xy}{ax - a^2} = \frac{(ax - ay) - (x^2 - xy)}{a(x - a)} = \frac{a(x - y) - x(x - y)}{a(x - a)} = \frac{(x - y)(a - x)}{a(x - a)} = \frac{(y - x)(x - a)}{a(x - a)} = \frac{y - x}{a}$
$\frac{b^2 - 2b + 1}{c - bc + a - ab} = \frac{(b - 1)^2}{(c - bc) + (a - ab)} = \frac{(b - 1)^2}{c(1 - b) + a(1 - b)} = \frac{(1 - b)^2}{(1 - b)(c + a)} = \frac{1 - b}{a + c}$
Пожауйста, оцените решение
