( ∗ ) x > 0 (*) x > 0 ( ∗ ) x > 0 log 3 ( 27 x 6 ) = 3 + 6 log 3 x \log_3(27x^6) = 3 + 6\log_3x log 3 ( 27 x 6 ) = 3 + 6 log 3 x t = log 3 x t = \log_3x t = log 3 x 1 + 6 t − 3 + 5 t 2 − 6 t − 3 + 12 ≥ 0 1 + \frac{6}{t - 3} + \frac{5}{t^2 - 6t - 3 + 12} \geq 0 1 + t − 3 6 + t 2 − 6 t − 3 + 12 5 ≥ 0 1 + 6 t − 3 + 5 t 2 − 6 t + 9 ≥ 0 1 + \frac{6}{t - 3} + \frac{5}{t^2 - 6t + 9} \geq 0 1 + t − 3 6 + t 2 − 6 t + 9 5 ≥ 0 1 + 6 t − 3 + 5 ( t − 3 ) 2 ≥ 0 1 + \frac{6}{t - 3} + \frac{5}{(t - 3)^2} \geq 0 1 + t − 3 6 + ( t − 3 ) 2 5 ≥ 0
Приведем к общему знаменателю ( t − 3 ) 2 (t-3)^2 ( t − 3 ) 2 ( t − 3 ) 2 + 6 ( t − 3 ) + 5 ( t − 3 ) 2 ≥ 0 \frac{(t-3)^2 + 6(t-3) + 5}{(t-3)^2} \geq 0 ( t − 3 ) 2 ( t − 3 ) 2 + 6 ( t − 3 ) + 5 ≥ 0 t 2 − 6 t + 9 + 6 t − 18 + 5 ( t − 3 ) 2 ≥ 0 \frac{t^2 - 6t + 9 + 6t - 18 + 5}{(t-3)^2} \geq 0 ( t − 3 ) 2 t 2 − 6 t + 9 + 6 t − 18 + 5 ≥ 0 t 2 − 4 ( t − 3 ) 2 ≥ 0 \frac{t^2 - 4}{(t-3)^2} \geq 0 ( t − 3 ) 2 t 2 − 4 ≥ 0 ( t − 2 ) ( t + 2 ) ( t − 3 ) 2 ≥ 0 \frac{(t-2)(t+2)}{(t-3)^2} \geq 0 ( t − 3 ) 2 ( t − 2 ) ( t + 2 ) ≥ 0
Получаем совокупность:
[ t ≤ − 2 t ≥ 2 t ≠ 3 ⇔ [ log 3 x ≤ − 2 log 3 x ≥ 2 log 3 x ≠ 3 ⇔ [ log 3 x ≤ log 3 1 9 log 3 x ≥ log 3 9 log 3 x ≠ log 3 27 \left[\begin{aligned}
&t \leq -2\\
& t \geq 2\\
&t \neq 3
\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}
&\log_3x \leq -2\\
& \log_3x \geq 2\\
&\log_3x \neq 3
\end{aligned}\right. \Leftrightarrow \left[\begin{aligned}
&\log_3x \leq \log_3\frac{1}{9}\\
& \log_3x \geq \log_39\\
&\log_3x \neq \log_327
\end{aligned}\right. t ≤ − 2 t ≥ 2 t = 3 ⇔ log 3 x ≤ − 2 log 3 x ≥ 2 log 3 x = 3 ⇔ log 3 x ≤ log 3 9 1 log 3 x ≥ log 3 9 log 3 x = log 3 27
Учитывая, что основания логарифмов 3 > 1 3 > 1 3 > 1 ( ∗ ) (*) ( ∗ ) { x > 0 [ x ≤ 1 9 x ≥ 9 x ≠ 27 \begin{cases}
x > 0\\
\left[\begin{aligned}
&x \leq \frac{1}{9}\\
& x \geq 9\\
&x \neq 27
\end{aligned}\right.
\end{cases} ⎩ ⎨ ⎧ x > 0 x ≤ 9 1 x ≥ 9 x = 27
Ответ: x ∈ ( 0 ; 1 9 ] ∪ [ 9 ; 27 ) ∪ ( 27 ; + ∞ ) x \in (0; \tfrac{1}{9}] \cup [9; 27) \cup (27; +\infty) x ∈ ( 0 ; 9 1 ] ∪ [ 9 ; 27 ) ∪ ( 27 ; + ∞ )
