Найдите значение выражения 10. \frac{x^{2}y-xy^{2}}{5(3y-x)} + \frac{2(x-3y)}{x^{2}-y^{2}} при х=-\frac{1}{7} и у=-14

Преобразуем выражение: $$\frac{x^{2}y-xy^{2}}{5(3y-x)} + \frac{2(x-3y)}{x^{2}-y^{2}} = \frac{xy(x-y)}{5(3y-x)} + \frac{2(x-3y)}{(x-y)(x+y)} = \frac{-xy(y-x)}{5(3y-x)} + \frac{2(x-3y)}{(x-y)(x+y)}$$ Приведём к общему знаменателю: $$\frac{-xy(y-x)(x-y)(x+y) + 10(x-3y)5(3y-x)}{5(3y-x)(x-y)(x+y)}$$ Теперь подставим значения $$x = -\frac{1}{7}$$ и $$y = -14$$: $$\frac{-\frac{-1}{7} \cdot (-14)(-\frac{1}{7} - (-14))}{5(3 \cdot (-14) - (-\frac{1}{7}))} + \frac{2(-\frac{1}{7} - 3 \cdot (-14))}{(-\frac{1}{7} - (-14))(-\frac{1}{7} + (-14))} = \frac{2(-\frac{1}{7} + 42)}{(-\frac{1}{7} + 14)(-\frac{1}{7} - 14)}$$ $$\frac{x^{2}y-xy^{2}}{5(3y-x)} + \frac{2(x-3y)}{(x-y)(x+y)}=\frac{xy(x-y)}{5(3y-x)} + \frac{2(x-3y)}{(x-y)(x+y)} = \frac{xy(x-y)}{-5(x-3y)} + \frac{2(x-3y)}{(x-y)(x+y)}$$ Если $$x=-\frac{1}{7}, y=-14$$, то $$\frac{-\frac{1}{7}*(-14)(-\frac{1}{7}+14)}{-5(-\frac{1}{7}-3*(-14))}+ \frac{2(-\frac{1}{7}-3*(-14))}{(-\frac{1}{7}+14)(-\frac{1}{7}-14)}=\frac{2*(-\frac{1}{7}+14)}{(-\frac{1}{7}+14)(-\frac{1}{7}-14)}=\frac{2}{(-\frac{1}{7}-14)}=\frac{2}{\frac{-1-98}{7}}=\frac{2}{\frac{-99}{7}}=\frac{14}{-99} \approx -0.14$$ Ответ: -0.8