Question

Решите неравенство:
$\frac{x + 3}{x} \geqslant {(x + 3)}^{2}$
​

Answer 1

(x + 3)/x ≥ (x + 3)²

(x + 3)² - (x + 3)/x ≤ 0

(x(x + 3)² - (x + 3))/x ≤ 0

(x + 3)(x² + 3x - 1)/x ≤ 0

x² + 3x - 1 = 0

D = 9 + 4 = 13

x12 = (-3 +- √13)/2

√13 ≈ 3.61

x1 = (-3 - √13)/2 = (-3 - 3.61)/2 < -3

x2 = (-3 + √13)/2 = (-3 + 3.61)/2 > 0

(x + 3)(x - x1)(x - x2)/x ≤ 0

Метод интервалов

+++++++++++[x1] ----------- [-3] +++++++++ (0) --------------- [x2] ++++++++++

x∈ [x1,-3] U (0, x2]

х1 и х2 даем обратно значения

x∈[ (-3-√13)/2, -3] U (0, (-3+√13)/2]