Решение №4197 Решите неравенство x^3+x^2-(4x^2-3x+6)/(x-2)

Решите неравенство $x^{3}+x^{2}–\frac{4x^{2}–3x+6}{x–2}\le 3$ .

Источник: statgrad

Решение:

x^{3}+x^{2}–\frac{4x^{2}–3x+6}{x–2}\le 3\\x^{3}+x^{2}–\frac{4x^{2}–3x+6}{x–2}- 3\le 0\\\frac{x^{3}\cdot (x–2)+x^{2}\cdot (x–2)-(4x^{2}–3x+6)\cdot 1-3\cdot (x–2)}{x–2}\le 0\\\frac{x^{4}–2x^{3}+x^{3}–2x^{2}-4x^{2}+3x-6-3x+6}{x–2}\le 0\\\frac{x^{4}–x^{3}–6x^{2}}{x–2}\le 0\\\frac{x^{2}\cdot (x^{2}–x–6)}{x–2}\le 0

Метод интервалов:

x – 2 ≠ 0
x ≠ 2

x² = 0
x₁ = 0

x² – x – 6 = 0
D = (–1)² – 4·1·(–6) = 1 + 24 = 25 = 5²
$x_{2}=\frac{1+5}{2\cdot 1}=\frac{6}{2}=3\\x_{3}=\frac{1-5}{2\cdot 1}=\frac{-4}{2}=-2$

x ∈ (–∞; –2] ∪ {0} ∪ (2; 3]

Ответ: (–∞; –2] ∪ {0} ∪ (2; 3].