Найдите значение выражения:
1) $\frac{1}{a^2 - ab} : \frac{b}{b^2 - a^2}$, если $a = 2\frac{1}{3}, b = -\frac{3}{7}$;
2) $\frac{a^2 + 4ab + 4b^2}{a^2 - 9b^2} : \frac{3a + 6b}{2a - 6b}$, если a = 4, b = −5.
$\frac{1}{a^2 - ab} : \frac{b}{b^2 - a^2} = \frac{1}{a(a - b)} : \frac{b}{(b - a)(b + a)} = \frac{1}{a(a - b)} : (-\frac{b}{(a - b)(a + b)}) = \frac{1}{a(a - b)} * (-\frac{(a - b)(a + b)}{b}) = \frac{1}{a} * (-\frac{a + b}{b}) = -\frac{a + b}{ab}$
при $a = 2\frac{1}{3}, b = -\frac{3}{7}$:
$-\frac{2\frac{1}{3} - \frac{3}{7}}{2\frac{1}{3} * (-\frac{3}{7})} = \frac{\frac{7}{3} - \frac{3}{7}}{\frac{7}{3} * \frac{3}{7}} = \frac{\frac{49 - 9}{21}}{\frac{1}{1} * \frac{1}{1}} = \frac{40}{21} = 1\frac{19}{21}$
$\frac{a^2 + 4ab + 4b^2}{a^2 - 9b^2} : \frac{3a + 6b}{2a - 6b} = \frac{(a + 2b)^2}{(a - 3b)(a + 3b)} : \frac{3(a + 2b)}{2(a - 3b)} = \frac{(a + 2b)^2}{(a - 3b)(a + 3b)} * \frac{2(a - 3b)}{3(a + 2b)} = \frac{a + 2b}{a + 3b} * \frac{2}{3} = \frac{2(a + 2b)}{3(a + 3b)}$
при a = 4, b = −5:
$\frac{2(a + 2b)}{3(a + 3b)} = \frac{2(4 + 2 * (-5))}{3(4 + 3 * (-5))} = \frac{2(4 - 10)}{3(4 - 15)} = \frac{2 * (-6)}{3 * (-11)} = \frac{2 * (-2)}{-11} = \frac{-4}{-11} = \frac{4}{11}$
Пожауйста, оцените решение
