5.86. Не вычисляя корней уравнения 2x² - 5x-4 = 0, найдите: 1 2 a) x² + x²; 6) x₁x²+x₂x₁; B) + ; ) x+x.

Дано уравнение: 2x² - 5x - 4 = 0. Разделим уравнение на 2, чтобы привести его к виду x² + px + q = 0: x² - \(\frac{5}{2}\)x - 2 = 0. По теореме Виета: x₁ + x₂ = \(\frac{5}{2}\) x₁x₂ = -2 a) \(\frac{1}{x_1^2} + \frac{1}{x_2^2}\) \(\frac{1}{x_1^2} + \frac{1}{x_2^2} = \frac{x_1^2 + x_2^2}{(x_1x_2)^2} = \frac{(x_1 + x_2)^2 - 2x_1x_2}{(x_1x_2)^2}\) Подставляем значения: \(\frac{(\frac{5}{2})^2 - 2(-2)}{(-2)^2} = \frac{\frac{25}{4} + 4}{4} = \frac{\frac{25 + 16}{4}}{4} = \frac{41}{16}\) б) x₁x₂⁴ + x₂x₁⁴ x₁x₂⁴ + x₂x₁⁴ = x₁x₂(x₁³ + x₂³) = x₁x₂((x₁ + x₂)^3 - 3x₁x₂(x₁ + x₂)) Подставляем значения: -2((\(\frac{5}{2}\))^3 - 3(-2)(\(\frac{5}{2}\))) = -2(\(\frac{125}{8}\) + 15) = -2(\(\frac{125 + 120}{8}\)) = -2(\(\frac{245}{8}\)) = -\(\frac{245}{4}\) в) \(\frac{x_1}{x_2^3} + \frac{x_2}{x_1^3}\) \(\frac{x_1}{x_2^3} + \frac{x_2}{x_1^3} = \frac{x_1^4 + x_2^4}{(x_1x_2)^3}\) Найдем x₁⁴ + x₂⁴ : x₁⁴ + x₂⁴ = (x₁² + x₂²)² - 2(x₁x₂)² = ((x₁ + x₂)² - 2x₁x₂)² - 2(x₁x₂)² = ((\(\frac{5}{2}\))^2 - 2(-2))² - 2(-2)² = (\(\frac{25}{4}\) + 4)² - 2(4) = (\(\frac{41}{4}\))² - 8 = \(\frac{1681}{16}\) - 8 = \(\frac{1681 - 128}{16}\) = \(\frac{1553}{16}\) Подставляем значения: \(\frac{\frac{1553}{16}}{(-2)^3} = \frac{\frac{1553}{16}}{-8} = -\frac{1553}{128}\) г) x₁⁵ + x₂⁵ x₁⁵ + x₂⁵ = (x₁² + x₂²)(x₁³ + x₂³) - x₁²x₂²(x₁ + x₂) x₁² + x₂² = (x₁ + x₂)² - 2x₁x₂ = (\(\frac{5}{2}\))² - 2(-2) = \(\frac{25}{4}\) + 4 = \(\frac{41}{4}\) x₁³ + x₂³ = (x₁ + x₂)^3 - 3x₁x₂(x₁ + x₂) = (\(\frac{5}{2}\))^3 - 3(-2)(\(\frac{5}{2}\)) = \(\frac{125}{8}\) + 15 = \(\frac{245}{8}\) x₁⁵ + x₂⁵ = (\(\frac{41}{4}\))(\(\frac{245}{8}\)) - (-2)²(\(\frac{5}{2}\)) = \(\frac{10045}{32}\) - 10 = \(\frac{10045 - 320}{32}\) = \(\frac{9725}{32}\)