1) 6 x - 10,2 = 4 x = - 2,2 x/0,2
2 x = 8 X = 4
2) 15 - ( 3 x - 3) = 5 - 4 x - 3 x + 4 x = 5 - 15 - 3 x = - 13
3) 2 x - 1 + 1 = 9
2 x = 9 + 1 - 1
2 x = 9
X = 4, 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)
А) (а+2)³=а³+3а² *2+3а*4+2³=а³+6а²+12а+8
б) (х-5)³=х³-3х² *5+3х*25-5³=х³-15х²+75х-125
а) 27х³+0,001=(3х)³+(0,1)³=(3х+0,1)(9х²-0,3х+0,01)
б) 125у³+m⁶=(5y)³+(m²)³=(5y+m²)(25y²-5m²y+m⁴)
в) 343а⁹-216=(7а³)³ - 6³=(7а³-6)(49а⁶+42а³+36)
A)sin2*75=sin150=1/2
b)sin π/4 = √2/2
c) 1/2 sin30 = 1/2 * 1/2=1/4
d)cos² π/8+sin² π/8 +2sin π/8cosπ/8 = 1+sinπ/4 = 1+ √2/2
e)sin² π/12 +cos²π/12 -2sin π/12cos<span>π/12= 1-sin </span>π/6 = 1-1/2=1/2
f)cos2*75 = cos 150=1/2
g)cos π/4 = √2/2
h)1-cos² π/12 - cos² π/12 = sin² π/12 - cos ² π/12 = -cos π/6= -√3/2