Выполните действия:
а) $(\frac{x}{y^2} - \frac{1}{x}) : (\frac{1}{y} + \frac{1}{x})$;
б) $(\frac{a}{m^2} + \frac{a^2}{m^3}) : (\frac{m^2}{a^2} + \frac{m}{a})$;
в) $\frac{ab + b^2}{3} : \frac{b^3}{3a} + \frac{a + b}{b}$;
г) $\frac{x - y}{x} - \frac{5y}{x^2} * \frac{x^2 - xy}{5y}$.
$(\frac{x}{y^2} - \frac{1}{x}) : (\frac{1}{y} + \frac{1}{x}) = \frac{x^2 - y^2}{xy^2} : \frac{x + y}{xy} = \frac{(x - y)(x + y)}{xy^2} * \frac{xy}{x + y} = \frac{x - y}{y} * \frac{1}{1} = \frac{x - y}{y}$
$(\frac{a}{m^2} + \frac{a^2}{m^3}) : (\frac{m^2}{a^2} + \frac{m}{a}) = \frac{a^2 + am}{m^3} : \frac{m^2 + am}{a^2} = \frac{a(a + m)}{m^3} * \frac{a^2}{m(m + a)} = \frac{a}{m^3} * \frac{a^2}{m} = \frac{a^3}{m^4}$
$\frac{ab + b^2}{3} : \frac{b^3}{3a} + \frac{a + b}{b} = \frac{b(a + b)}{3} * \frac{3a}{b^3} + \frac{a + b}{b} = \frac{a + b}{1} * \frac{a}{b^2} + \frac{a + b}{b} = \frac{a(a + b)}{b^2} + \frac{a + b}{b} = \frac{a(a + b) + b(a + b)}{b^2} = \frac{(a + b)(a + b)}{b^2} = \frac{(a + b)^2}{b^2}$
$\frac{x - y}{x} - \frac{5y}{x^2} * \frac{x^2 - xy}{5y} = \frac{x - y}{x} - \frac{5y}{x^2} * \frac{x(x - y)}{5y} = \frac{x - y}{x} - \frac{1}{x} * \frac{x - y}{1} = \frac{x - y}{x} - \frac{x - y}{x} = 0$
Пожауйста, оцените решение
