1-(8а-3)^2
(1-(8а-3))×(1+(8а-3))
(1-8а+3)×(1+8а-3)
(4-8а)×(-2+8а)
4(1-2а)×(-2(1-4а))
-4×2(1-2а)×(1-4а)
-8(1-2а)×(1-4а)
(2a+b)²(b-2a) = (4а²+4ab+b²)(b-2a) = 4a²b-8a³+4ab²-8a²b+b³-2ab² = -4a²b-8a³+2ab²+b³
А)у=х^2+4x-2x-8
y=x^2+2x-8
Ищем дискрименант:
D=36=6^2
x1=2 x2= -4
б)y=-х(х+5)
x=0 x=-5
13(5x-1)-15(4x+2)<0
65-13-60x-30<0
5x-43<0
5x<43
x<43\5
x<8,6
![\displaystyle \dfrac{1!}{1^1}+\dfrac{2!}{2^2}+...+\dfrac{n!}{n^n}=\sum^{\infty}_{n=1}\dfrac{n!}{n^n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cdfrac%7B1%21%7D%7B1%5E1%7D%2B%5Cdfrac%7B2%21%7D%7B2%5E2%7D%2B...%2B%5Cdfrac%7Bn%21%7D%7Bn%5En%7D%3D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D1%7D%5Cdfrac%7Bn%21%7D%7Bn%5En%7D)
Здесь общий член ряда
. Тогда по признаку Коши
![\displaystyle \lim_{n \to \infty} \sqrt[n]{a_n}= \lim_{n \to \infty}\dfrac{\sqrt[n]{n!}}{n}~~ =\boxed{=}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csqrt%5Bn%5D%7Ba_n%7D%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cdfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%7D%7Bn%7D~~%20%3D%5Cboxed%7B%3D%7D)
По формуле Стирлинга
, мы получим
![\boxed{=}~\displaystyle \lim_{n \to \infty}\dfrac{(2\pi n)^{1/2n}\cdot\frac{n}{e}}{n}=\dfrac{1}{e}<1](https://tex.z-dn.net/?f=%5Cboxed%7B%3D%7D~%5Cdisplaystyle%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cdfrac%7B%282%5Cpi%20n%29%5E%7B1%2F2n%7D%5Ccdot%5Cfrac%7Bn%7D%7Be%7D%7D%7Bn%7D%3D%5Cdfrac%7B1%7D%7Be%7D%3C1)
Данный ряд сходится.