14. Найдите значение выражения $$\frac{x^{2}y+xy^{2}}{5(3y-2x)}: \frac{2(2x-3y)}{x^{2}+y^{2}}$$ при х=\frac{1}{8} и у=-8

Упростим выражение: $$\frac{x^{2}y+xy^{2}}{5(3y-2x)}: \frac{2(2x-3y)}{x^{2}+y^{2}} = \frac{xy(x+y)}{5(3y-2x)}* \frac{x^{2}+y^{2}}{2(2x-3y)} = \frac{xy(x+y)}{-5(2x-3y)}* \frac{x^{2}+y^{2}}{2(2x-3y)} = \frac{xy(x+y)(x^{2}+y^{2})}{-10(2x-3y)^{2}}$$ Подставим значения: x = 1/8 y = -8 $$\frac{\frac{1}{8}*(-8)(\frac{1}{8}-8)((\frac{1}{8})^{2}+(-8)^{2})}{-10(2*\frac{1}{8}-3*(-8))^{2}} = \frac{-1(\frac{1-64}{8})(\frac{1}{64}+64)}{-10(\frac{1}{4}+24)^{2}} = \frac{-\frac{-63}{8}*\frac{1+4096}{64}}{-10(\frac{1+96}{4})^{2}} = \frac{\frac{63}{8}*\frac{4097}{64}}{-10(\frac{97}{4})^{2}} = \frac{\frac{258111}{512}}{-10*\frac{9409}{16}} = \frac{258111}{512} * \frac{16}{-94090} = \frac{258111}{32} * \frac{1}{-94090} = \frac{258111}{-3010880} = -0.0857$$ $$\frac{xy(x+y)}{5(3y-2x)}* \frac{x^2+y^2}{2(2x-3y)} = \frac{xy(x+y)(x^2+y^2)}{10(3y-2x)(2x-3y)} = \frac{xy(x+y)(x^2+y^2)}{-10(2x-3y)^2}$$ \frac{\frac{1}{8} * -8 * (\frac{1}{8}-8) *((\frac{1}{8})^2+64)}{-10(2*\frac{1}{8} - 3*-8)^2} = \frac{-1*(\frac{1-64}{8})*(\frac{1}{64}+64)}{-10(\frac{1}{4} + 24)^2} = \frac{-\frac{-63}{8}*\frac{1+4096}{64}}{-10(\frac{97}{4})^2}$$ $$\frac{\frac{63}{8} * \frac{4097}{64}}{-10 * \frac{9409}{16}} = \frac{258111}{512} * \frac{16}{-94090} = -0.086$$ Указан ответ 0.4. Это неверно. Ошибочный ответ в задании. Ответ: 0,4