( ∗ ) { 3 ∣ 4 x + 1 ∣ ≠ 1 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 > 0 ⇔ { 3 ∣ 4 x + 1 ∣ ≠ 3 0 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 > 0 ⇔ { ∣ 4 x + 1 ∣ ≠ 0 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 > 0 (*)\begin{cases}
3^{|4x + 1|} \neq 1\\
\tfrac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4} > 0
\end{cases} \Leftrightarrow \begin{cases}
3^{|4x + 1|} \neq 3^0\\
3^{2x + 1} - 2\cdot3^{x + 1} + 3 > 0
\end{cases} \Leftrightarrow \begin{cases}
|4x + 1| \neq 0\\
3^{2x + 1} - 2\cdot3^{x + 1} + 3 > 0
\end{cases} ( ∗ ) { 3 ∣4 x + 1∣ = 1 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 > 0 ⇔ { 3 ∣4 x + 1∣ = 3 0 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 > 0 ⇔ { ∣4 x + 1∣ = 0 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 > 0 { x ≠ − 1 4 3 2 x − 2 ⋅ 3 x + 1 > 0 ⇔ { x ≠ − 1 4 ( 3 x − 1 ) 2 > 0 ⇔ { x ≠ − 1 4 x ≠ 0 \begin{cases}
x \neq -\tfrac{1}{4}\\
3^{2x} - 2\cdot3^{x} + 1 > 0
\end{cases} \Leftrightarrow \begin{cases}
x \neq -\tfrac{1}{4}\\
(3^x - 1)^2> 0
\end{cases} \Leftrightarrow \begin{cases}
x \neq -\tfrac{1}{4}\\
x \neq 0
\end{cases} { x = − 4 1 3 2 x − 2 ⋅ 3 x + 1 > 0 ⇔ { x = − 4 1 ( 3 x − 1 ) 2 > 0 ⇔ { x = − 4 1 x = 0
Вернёмся к неравенству:
log 3 ∣ 4 x + 1 ∣ ( 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 ) ≤ x ∣ 4 x + 1 ∣ \log_{3^{|4x + 1|}}\left(\frac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4}\right) \leq \frac{x}{|4x + 1|} log 3 ∣4 x + 1∣ ( 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 ) ≤ ∣4 x + 1∣ x log a b c = 1 b log a c \log_{a^b}c = \frac{1}{b}\log_ac log a b c = b 1 log a c
Тогда:
1 ∣ 4 x + 1 ∣ log 3 ( 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 ) ≤ x ∣ 4 x + 1 ∣ \frac{1}{|4x + 1|}\log_3\left(\frac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4}\right) \leq \frac{x}{|4x + 1|} ∣4 x + 1∣ 1 log 3 ( 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 ) ≤ ∣4 x + 1∣ x ( ∗ ) (*) ( ∗ ) ∣ 4 x + 1 ∣ |4x + 1| ∣4 x + 1∣ log 3 ( 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 ) ≤ x ⇔ log 3 ( 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 ) ≤ log 3 3 x \log_3\left(\frac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4}\right) \leq x \Leftrightarrow \log_3\left(\frac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4}\right) \leq \log_33^x log 3 ( 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 ) ≤ x ⇔ log 3 ( 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 ) ≤ log 3 3 x 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 4 ≤ 3 x ⇔ 3 ⋅ 3 2 x − 6 ⋅ 3 x + 3 − 4 ⋅ 3 x ≤ 0 ⇔ \frac{3^{2x + 1} - 2\cdot3^{x + 1} + 3}{4} \leq 3^x \Leftrightarrow 3\cdot3^{2x} - 6\cdot3^x+3 - 4\cdot3^x \leq 0 \Leftrightarrow 4 3 2 x + 1 − 2 ⋅ 3 x + 1 + 3 ≤ 3 x ⇔ 3 ⋅ 3 2 x − 6 ⋅ 3 x + 3 − 4 ⋅ 3 x ≤ 0 ⇔ ⇔ 3 ⋅ 3 2 x − 10 ⋅ 3 x + 3 ≤ 0 \Leftrightarrow 3\cdot3^{2x} - 10\cdot3^x+3 \leq 0 ⇔ 3 ⋅ 3 2 x − 10 ⋅ 3 x + 3 ≤ 0 3 x = t > 0 3^x = t > 0 3 x = t > 0 3 t 2 − 10 t + 3 ≤ 0 ⇔ ( 3 t − 1 ) ( t − 3 ) ≤ 0 3t^2 - 10t + 3 \leq 0 \Leftrightarrow (3t - 1)(t - 3) \leq 0 3 t 2 − 10 t + 3 ≤ 0 ⇔ ( 3 t − 1 ) ( t − 3 ) ≤ 0
t ∈ [ 1 3 ; 3 ] t \in \left[\frac{1}{3}; 3\right] t ∈ [ 3 1 ; 3 ] { t ≥ 1 3 t ≤ 3 ⇔ { 3 x ≥ 1 3 3 x ≤ 3 ⇔ { 3 x ≥ 3 − 1 3 x ≤ 3 1 ⇔ { x ≥ − 1 x ≤ 1 \begin{cases}
t \geq \tfrac{1}{3}\\
t \leq 3
\end{cases} \Leftrightarrow \begin{cases}
3^x \geq \tfrac{1}{3}\\
3^x \leq 3
\end{cases} \Leftrightarrow \begin{cases}
3^x \geq 3^{-1}\\
3^x \leq 3^1
\end{cases} \Leftrightarrow \begin{cases}
x \geq -1\\
x \leq 1
\end{cases} { t ≥ 3 1 t ≤ 3 ⇔ { 3 x ≥ 3 1 3 x ≤ 3 ⇔ { 3 x ≥ 3 − 1 3 x ≤ 3 1 ⇔ { x ≥ − 1 x ≤ 1 ( ∗ ) (*) ( ∗ ) { x ≥ − 1 x ≤ 1 x ≠ − 1 4 x ≠ 0 ⇒ x ∈ [ − 1 ; − 1 4 ) ∪ ( − 1 4 ; 0 ) ∪ ( 0 ; 1 ] \begin{cases}
x \geq -1\\
x \leq 1\\
x \neq -\tfrac{1}{4}\\
x \neq 0
\end{cases} \Rightarrow x \in \left[-1; -\frac{1}{4}\right) \cup \left(-\frac{1}{4}; 0\right) \cup (0;1] ⎩ ⎨ ⎧ x ≥ − 1 x ≤ 1 x = − 4 1 x = 0 ⇒ x ∈ [ − 1 ; − 4 1 ) ∪ ( − 4 1 ; 0 ) ∪ ( 0 ; 1 ]
Ответ: x ∈ [ − 1 ; − 1 4 ) ∪ ( − 1 4 ; 0 ) ∪ ( 0 ; 1 ] x \in \left[-1; -\frac{1}{4}\right) \cup \left(-\frac{1}{4}; 0\right) \cup (0;1] x ∈ [ − 1 ; − 4 1 ) ∪ ( − 4 1 ; 0 ) ∪ ( 0 ; 1 ]
