Вопрос школьника по предмету Алгебра
logx-3(x^2-4x+3)>=1 Решите неравенство
Ответ учителя по предмету Алгебра
logₓ₋₃(x^2-4x+3) ≥
1
если основание > 1, то данная логарифмическая функция возрастающая и наоборот.
Так что в нашем примере возможны 2 варианта
а) х — 3 > 1 б) 0 < х — 3 <1
х > 4 3 < x < 4
теперь решаем, учитывая ОДЗ
х² — 4х +3 > 0 х² — 4х +3 > 0
x² — 4x + 3 > x — 3 x² — 4x + 3 < x — 3
решаем обе эти системы
х² -4х + 3 корни 1 и 3
х² -5х +6 корни 2 и 3
-∞ 1 2 3 4 +∞
IIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIII
Ответ: (4;+∞ )
б)-∞ 1 2 3 4 +∞
IIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIII
IIIIIIIIIII
∅
Ответ: (4;+∞ )
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