5. Найдите значение выражения (x² - x⁵) : 2(x - 3y) * 1 / (5(3y - x)) при x = -1/7 и y = -14.

Решение:

1. Упростим выражение:
2. \( \frac{x^2 - x^5}{2(x - 3y)} \cdot \frac{1}{5(3y - x)} = \frac{x^2(1 - x^3)}{2(x - 3y)} \cdot \frac{1}{5(-(x - 3y))} = -\frac{x^2(1 - x^3)}{10(x - 3y)^2} \)
3. Подставим значения \( x = -\frac{1}{7} \) и \( y = -14 \):
4. \( x - 3y = -\frac{1}{7} - 3(-14) = -\frac{1}{7} + 42 = \frac{-1 + 294}{7} = \frac{293}{7} \)
5. \( (x - 3y)^2 = (\frac{293}{7})^2 = \frac{85849}{49} \)
6. \( x^2 = (-\frac{1}{7})^2 = \frac{1}{49} \)
7. \( 1 - x^3 = 1 - (-\frac{1}{7})^3 = 1 - (-\frac{1}{343}) = 1 + \frac{1}{343} = \frac{344}{343} \)
8. \( -\frac{x^2(1 - x^3)}{10(x - 3y)^2} = -\frac{\frac{1}{49} \cdot \frac{344}{343}}{10 \cdot \frac{85849}{49}} = -\frac{\frac{344}{49 \u0002 343}}{10 \u0002 \frac{85849}{49}} = -\frac{344}{49 \u0002 343} \cdot \frac{49}{10 \u0002 85849} = -\frac{344}{343 \u0002 10 \u0002 85849} \)
9. \( -\frac{344}{3430 \u0002 85849} = -\frac{344}{294252070} \approx -1.17 \cdot 10^{-6} \)

Ответ: $$\approx -1.17 \times 10^{-6}$$