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golovanzhenya
@golovanzhenya
August 2022
1
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Смотри вложение!!
Ребятя,срочно нужно!!!
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Verified answer
(log₃x)²-I(log₃x)I<2
заменим I(log₃x)I =t ≥0
т.к. (log₃x)² = I(log₃x)I² , тогда (log₃x)²-I(log₃x)I<2 t²-t <2
t²-t-2 <0
t ≥0 ⇔ t
≥0
t²-t-2 = 0 t1=-1 t2=2
+ - - +
-------------(-1)--//-----------0-------///-------2------------
i-----///------------------------------------
0≤t≤2 ⇔ I(log₃x)I≤2 ⇔ -2≤(log₃x)≤2 ⇔3∧(-2)≤x≤3∧2 ⇔ 1/9≤x≤9
ответ: 1/9≤x≤9
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Answers & Comments
Verified answer
(log₃x)²-I(log₃x)I<2заменим I(log₃x)I =t ≥0
т.к. (log₃x)² = I(log₃x)I² , тогда (log₃x)²-I(log₃x)I<2 t²-t <2 t²-t-2 <0
t ≥0 ⇔ t ≥0
t²-t-2 = 0 t1=-1 t2=2
+ - - +
-------------(-1)--//-----------0-------///-------2------------
i-----///------------------------------------
0≤t≤2 ⇔ I(log₃x)I≤2 ⇔ -2≤(log₃x)≤2 ⇔3∧(-2)≤x≤3∧2 ⇔ 1/9≤x≤9
ответ: 1/9≤x≤9