1) 9x-3
![x^{2}](https://tex.z-dn.net/?f=+x%5E%7B2%7D+)
< 0
3x (3-x) < 0
x=0 x=3
Ответ: (-бесконечность;0) и (3;+бесконечность)
2) 25-
![x^{2}](https://tex.z-dn.net/?f=+x%5E%7B2%7D+)
> 0
![x^{2}](https://tex.z-dn.net/?f=+x%5E%7B2%7D+)
< 25
x=
+- 5
Ответ: (-5;5)
Рассмотрим предел
![\displaystyle \lim_{x \to 0}\dfrac{(1+x)^p-1}{x}=\lim_{x \to 0}\dfrac{((1+x)^p-1)'}{(x)'}=\lim_{x \to 0}\dfrac{p(1+x)^{p-1}}{1}=p\\ \\ \\ \lim_{x \to 0}\dfrac{(1+x)^p-1}{x}=p~~~~\Rightarrow~~~ (1+x)^p-1~~\sim xp](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7B%281%2Bx%29%5Ep-1%7D%7Bx%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7B%28%281%2Bx%29%5Ep-1%29%27%7D%7B%28x%29%27%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7Bp%281%2Bx%29%5E%7Bp-1%7D%7D%7B1%7D%3Dp%5C%5C%20%5C%5C%20%5C%5C%20%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7B%281%2Bx%29%5Ep-1%7D%7Bx%7D%3Dp~~~~%5CRightarrow~~~%20%281%2Bx%29%5Ep-1~~%5Csim%20xp)
Второй способ доказательства (без Лопиталя)
![\displaystyle \lim_{x \to 0}\dfrac{(1+x)^p-1}{xp}=\lim_{x \to 0}\dfrac{e^{p\ln(1+x)}-1}{xp}=\lim_{x \to 0}\dfrac{e^{p\ln(1+x)-1}}{p\ln(1+x)}\cdot \\ \\ \cdot\dfrac{\ln(1+x)}{x}=\lim_{x \to 0}\dfrac{e^{p\ln(1+x)}-1}{p\ln(1+x)}\cdot 1=\left\{\begin{array}{ccc}p\ln(1+x)=t\\ \\ t\to 0\end{array}\right\}=\\ \\ \\ =\lim_{t \to 0}\dfrac{e^t-1}{t}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7B%281%2Bx%29%5Ep-1%7D%7Bxp%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7Be%5E%7Bp%5Cln%281%2Bx%29%7D-1%7D%7Bxp%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7Be%5E%7Bp%5Cln%281%2Bx%29-1%7D%7D%7Bp%5Cln%281%2Bx%29%7D%5Ccdot%20%5C%5C%20%5C%5C%20%5Ccdot%5Cdfrac%7B%5Cln%281%2Bx%29%7D%7Bx%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7Be%5E%7Bp%5Cln%281%2Bx%29%7D-1%7D%7Bp%5Cln%281%2Bx%29%7D%5Ccdot%201%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dp%5Cln%281%2Bx%29%3Dt%5C%5C%20%5C%5C%20t%5Cto%200%5Cend%7Barray%7D%5Cright%5C%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D%5Clim_%7Bt%20%5Cto%200%7D%5Cdfrac%7Be%5Et-1%7D%7Bt%7D%3D1)
Отсюда следует, что
при ![x\to 0](https://tex.z-dn.net/?f=x%5Cto%200)
2cos(x+п/3)=к3
соs( x+п/3)=k3/2
x+п/3=+-arccos k3/2+2пn
x=+-п/6-п/3+2пn
Из тождества
![1+tg^2 \alpha = \dfrac{1}{\cos^2\alpha }](https://tex.z-dn.net/?f=1%2Btg%5E2%20%5Calpha%20%3D%20%5Cdfrac%7B1%7D%7B%5Ccos%5E2%5Calpha%20%7D%20)
имеем, что
![tg^2\alpha = \dfrac{1}{\cos^2\alpha }-1= \dfrac{1}{(1/\sqrt{10})^2} -1=10-1=9](https://tex.z-dn.net/?f=tg%5E2%5Calpha%20%3D%20%5Cdfrac%7B1%7D%7B%5Ccos%5E2%5Calpha%20%7D-1%3D%20%5Cdfrac%7B1%7D%7B%281%2F%5Csqrt%7B10%7D%29%5E2%7D%20%20-1%3D10-1%3D9)
Поскольку
![( \frac{3 \pi }{2} ;2 \pi )](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B3%20%5Cpi%20%7D%7B2%7D%20%3B2%20%5Cpi%20%29)
IV четверть, то в этой четверти тангенс отрицателен.
- ОТВЕТ.