ОДЗ :
x² - 2x - 8 > 0
(x - 4)(x + 2) > 0
+ - +
_____________₀__________₀___________
- 2 4
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x ∈ ( - ∞ , - 2) ∪ (4 ; + ∞)
0\\\\log_{\frac{1}{3} }(x^{2}-2x-8)>-3\\\\x^{2}-2x-8<27\\\\x^{2}-2x-35<0\\\\x^{2}-2x-35=0\\\\D=(-2)^{2}-4*(-35)=4+140=144=12^{2}\\\\x_{1}=\frac{2+12}{2}=7\\\\x_{2}=\frac{2-12}{2} =-5" alt="log_{\frac{1}{3} }(x^{2}-2x-8)+3>0\\\\log_{\frac{1}{3} }(x^{2}-2x-8)>-3\\\\x^{2}-2x-8<27\\\\x^{2}-2x-35<0\\\\x^{2}-2x-35=0\\\\D=(-2)^{2}-4*(-35)=4+140=144=12^{2}\\\\x_{1}=\frac{2+12}{2}=7\\\\x_{2}=\frac{2-12}{2} =-5" align="absmiddle" class="latex-formula">
+ - +
______________₀_____________₀_____________
- 5 7
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x ∈ (- 5 , 7)
С учётом ОДЗ окончательный ответ :
x ∈ (- 5 ; - 2) ∪ (4 , 7)