(-x² - 36x)/(x² + x + 1)(x + 5) ≥ 0
x(x + 36) /(x² + x + 1)(x + 5) ≤ 0
x² + x + 1 > 0 (D = 1 - 4 < 0) можно опустить
x(x + 36) /(x + 5) ≤ 0
Метод интервалов
----------[-36] ++++++(-5) --------------[0] +++++++
x ∈(-∞, -36] U (-5,0]
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Verified answer
(-x² - 36x)/(x² + x + 1)(x + 5) ≥ 0
x(x + 36) /(x² + x + 1)(x + 5) ≤ 0
x² + x + 1 > 0 (D = 1 - 4 < 0) можно опустить
x(x + 36) /(x + 5) ≤ 0
Метод интервалов
----------[-36] ++++++(-5) --------------[0] +++++++
x ∈(-∞, -36] U (-5,0]