Применена формула сложения
ОДЗ :
x² - 2x - 8 > 0
(x - 4)(x + 2) > 0
+ - +
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- 2 4
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x ∈ ( - ∞ , - 2) ∪ (4 ; + ∞)
![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](https://tex.z-dn.net/?f=log_%7B%5Cfrac%7B1%7D%7B3%7D+%7D%28x%5E%7B2%7D-2x-8%29%2B3%3E0%5C%5C%5C%5Clog_%7B%5Cfrac%7B1%7D%7B3%7D+%7D%28x%5E%7B2%7D-2x-8%29%3E-3%5C%5C%5C%5Cx%5E%7B2%7D-2x-8%3C27%5C%5C%5C%5Cx%5E%7B2%7D-2x-35%3C0%5C%5C%5C%5Cx%5E%7B2%7D-2x-35%3D0%5C%5C%5C%5CD%3D%28-2%29%5E%7B2%7D-4%2A%28-35%29%3D4%2B140%3D144%3D12%5E%7B2%7D%5C%5C%5C%5Cx_%7B1%7D%3D%5Cfrac%7B2%2B12%7D%7B2%7D%3D7%5C%5C%5C%5Cx_%7B2%7D%3D%5Cfrac%7B2-12%7D%7B2%7D+%3D-5)
+ - +
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- 5 7
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x ∈ (- 5 , 7)
С учётом ОДЗ окончательный ответ :
x ∈ (- 5 ; - 2) ∪ (4 , 7)
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4х-7х+2=17
-3х=17-2
-3х=15 |:(-3)
х=-5
ответ:х=-5
Y=100/(1+5t)=100*(1+5t)⁻¹
y'=-100*5/(1+5t)² v=|y'(t)|=500/(1+5t)²