log (x) (1 - x) < 1
одз x > 0
x ≠ 1
x < 1
x ∈ (0,1)
log (x) (1 - x) < log(x) x ⇔
(x - 1)(1 - x - x) < 0
(x - 1)(1 - 2x) < 0
(x - 1)(2x - 1) > 0
++++++(1/2) ---------- (1) ++++++++
x ∈ (-∞ 1/2) U (1 +∞) пересекаем с одз
ответ x∈(0, 1/2)
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Verified answer
log (x) (1 - x) < 1
одз x > 0
x ≠ 1
x < 1
x ∈ (0,1)
log (x) (1 - x) < log(x) x ⇔
(x - 1)(1 - x - x) < 0
(x - 1)(1 - 2x) < 0
(x - 1)(2x - 1) > 0
++++++(1/2) ---------- (1) ++++++++
x ∈ (-∞ 1/2) U (1 +∞) пересекаем с одз
ответ x∈(0, 1/2)