Ответ:
пардоньте С не знаю как решить, спаибо было интересно
X^2 - 2x = x + 2 - x^2
x^2 - 2x - x - 2 + x^2 = 0
2x^2 - 3x - 2 = 0
D = 9 + 4*4 = 16 + 9 = 25
x1 = ( 3 + 5)/4 = 8/4 = 2
x2 = ( 3 - 5 )/4 = - 2/4 = - 1/2 = - 0,5
2)
(10 + 2√10) / √10 = √10*(√10 + 2) / √10 = √10 + 2,
3)
(х - 18√х + 81) / (х - 81) = (√х - 9) / (√х - 9)(√х + 9) = 1 / (√х + 9),
6 / (√11 - 3) = 6(√11 + 3) / (√11 - 3)(√11 + 3) =
= 6(√11 + 3) / (11 - 9) = 6(√11 + 3) / 2 = 3*(√11 + 3),
1)
√125х¹² = √(25*5*(х⁶)²) = 5х⁶√5,
2)
√(а⁷*(в¹¹)² = в¹¹*√а⁷,
2)
(√24 - √6) * √6 = √(24*6) - √6*√6 = √(4*6*6) - 6 =
= 2*6 - 6 = 12 - 6 = 6,
3)
(√6 - 1)² = (√6)² - 2*1*√6 + 1² = 6 - 2√6 + 1 = 7 - 2√6
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Последовательность возрастает, если
![d_{n+1}\ \textgreater \ d_n](https://tex.z-dn.net/?f=d_%7Bn%2B1%7D%5C+%5Ctextgreater+%5C+d_n)
Докажем, что
![\frac{8(n+1)}{(n+1)+1}\ \textgreater \ \frac{8n}{n+1}](https://tex.z-dn.net/?f=%5Cfrac%7B8%28n%2B1%29%7D%7B%28n%2B1%29%2B1%7D%5C+%5Ctextgreater+%5C+%5Cfrac%7B8n%7D%7Bn%2B1%7D+)
Рассмотрим разность
![\frac{8(n+1)}{(n+1)+1}- \frac{8n}{n+1}=\frac{8(n+1)(n+1)}{(n+1)(n+2)}- \frac{8n(n+2)}{(n+1)(n+2)}=\frac{8(n+1)(n+1)-8n(n+2)}{(n+1)(n+2)}= \\ \\ =\frac{8n^2+16n+8-8n^2-16n}{(n+1)(n+2)}=\frac{8}{(n+1)(n+2)}\ \textgreater \ 0](https://tex.z-dn.net/?f=%5Cfrac%7B8%28n%2B1%29%7D%7B%28n%2B1%29%2B1%7D-+%5Cfrac%7B8n%7D%7Bn%2B1%7D%3D%5Cfrac%7B8%28n%2B1%29%28n%2B1%29%7D%7B%28n%2B1%29%28n%2B2%29%7D-+%5Cfrac%7B8n%28n%2B2%29%7D%7B%28n%2B1%29%28n%2B2%29%7D%3D%5Cfrac%7B8%28n%2B1%29%28n%2B1%29-8n%28n%2B2%29%7D%7B%28n%2B1%29%28n%2B2%29%7D%3D+%5C%5C+%5C%5C+%3D%5Cfrac%7B8n%5E2%2B16n%2B8-8n%5E2-16n%7D%7B%28n%2B1%29%28n%2B2%29%7D%3D%5Cfrac%7B8%7D%7B%28n%2B1%29%28n%2B2%29%7D%5C+%5Ctextgreater+%5C+0)
Значит,
![\frac{8(n+1)}{(n+1)+1}\ \textgreater \ \frac{8n}{n+1}](https://tex.z-dn.net/?f=%5Cfrac%7B8%28n%2B1%29%7D%7B%28n%2B1%29%2B1%7D%5C+%5Ctextgreater+%5C+%5Cfrac%7B8n%7D%7Bn%2B1%7D+)