Task/26151217
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см приложение
решите неравенство - 2Log_(x/3) 27 ≥ Log_(3) 27x + 1 ;
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решение : - 2 Log_(x/3) 3³ ≥ Log_(3) 27+Log_(3) x + 1 ; - 6 Log_(x/3) 3 ≥ 3 + Log_(3) x + 1 ;
OДЗ : { x > 0 ; x/3 ≠1 ⇔ x ∈ ( 0 ; 3) ∪ (3 ;∞) .
* * Log_(x/3) 3 = 1/ Log_(3) x/3 =1/( Log_(3) x - Log_(3) 3) =1/ ( Log_(3) x - 1) * * *
- 6 / ( Log_(3) x - 1 ) ≥ Log_(3) x + 4 ;
Log_(3) x + 4 + 6 / ( Log_(3) x - 1 ) ≤ 0 ;
* * можно замена : t = Log_(3) x ; (t² +3t +2)/(t -1) ≤ 0⇔(t +2)(t +1)/(t -1) ≤ 0 * * ( Log²_(3) x +3Log_(3) x + 2 ) / ( Log_(3) x - 1 ) ≤ 0 ;
(Log_(3) x +2)(Log_(3) x +1) / ( Log_(3) x - 1 ) ≤ 0 ;
- + - +
Log_(3) x : //////////////// [ -2] ----------[ -1] /////////////// (1) ------------
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ответ: x ∈ ( 0 ; 1/9 ] ∪ [1/3 ; 3)
