Найдите значение выражения \(\frac{\text{x}^2+\text{x}^3}{5(\text{x}-3\text{y})} \cdot \frac{2(\text{y}-\text{x})}{(\text{x}^2-\text{y}^2)}\) при \(\text{x}=-3\) и \(\text{y}=\frac{1}{3}\).

Решение:

Упростим выражение:

\[ \frac{\text{x}^2(1+\text{x})}{5(\text{x}-3\text{y})} \cdot \frac{2(\text{y}-\text{x})}{(\text{x}-\text{y})(\text{x}+\text{y})} = \frac{\text{x}^2(1+\text{x})}{5(\text{x}-3\text{y})} \cdot \frac{-2(\text{x}-\text{y})}{(\text{x}-\text{y})(\text{x}+\text{y})} = \frac{-2\text{x}^2(1+\text{x})}{5(\text{x}-3\text{y})(\text{x}+\text{y})} \]

Подставим значения \(\text{x}=-3\) и \(\text{y}=\frac{1}{3}\):

\[ \frac{-2(-3)^2(1+(-3))}{5(-3-3(\frac{1}{3}))(-3+\frac{1}{3})} = \frac{-2(9)(-2)}{5(-3-1)(-3+\frac{1}{3})} = \frac{36}{5(-4)(-\frac{8}{3})} = \frac{36}{5(\frac{32}{3})} = \frac{36}{\frac{160}{3}} = 36 \cdot \frac{3}{160} = \frac{108}{160} = \frac{27}{40} \]

Ответ: \(\frac{27}{40}\)