3) так как перемножив первую и вторую скобки мы получи квадратное уравнение с D=0,а значит,с двумя одинаковыми корнями,которые и равны 2
![x^{4}+\frac{x^{4}}{1+x^{4}}+\frac{x^{4} }{(1+x^{4})^{2}}+\frac{x^{4}}{(1+x^{4})^{3}}+...}=x^{4} (1+\frac{1}{1+x^{4}}+\frac{1}{(1+x^{4})^{2}}+\frac{1}{(1+x^{4})^{3}}+...)](https://tex.z-dn.net/?f=x%5E%7B4%7D%2B%5Cfrac%7Bx%5E%7B4%7D%7D%7B1%2Bx%5E%7B4%7D%7D%2B%5Cfrac%7Bx%5E%7B4%7D%20%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B2%7D%7D%2B%5Cfrac%7Bx%5E%7B4%7D%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B3%7D%7D%2B...%7D%3Dx%5E%7B4%7D%20%281%2B%5Cfrac%7B1%7D%7B1%2Bx%5E%7B4%7D%7D%2B%5Cfrac%7B1%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B3%7D%7D%2B...%29)
То что в скобках - это бесконечно убывающая геометрическая прогрессия в которой b₁ = 1 , а q = 1/(1+x⁴).
Найдём сумму этой прогрессии :
![S=\frac{b_{1} }{1-q} =\frac{1}{1-\frac{1}{1+x^{4}}}=\frac{1}{\frac{1+x^{4}-1 }{1+x^{4}}}=\frac{1}{\frac{x^{4} }{1+x^{4}}}=\frac{1+x^{4} }{x^{4}}](https://tex.z-dn.net/?f=S%3D%5Cfrac%7Bb_%7B1%7D%20%7D%7B1-q%7D%20%3D%5Cfrac%7B1%7D%7B1-%5Cfrac%7B1%7D%7B1%2Bx%5E%7B4%7D%7D%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B1%2Bx%5E%7B4%7D-1%20%7D%7B1%2Bx%5E%7B4%7D%7D%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7Bx%5E%7B4%7D%20%7D%7B1%2Bx%5E%7B4%7D%7D%7D%3D%5Cfrac%7B1%2Bx%5E%7B4%7D%20%7D%7Bx%5E%7B4%7D%7D)
Следовательно :
![x^{4}+\frac{x^{4} }{1+x^{4}}+\frac{x^{4}}{(1+x^{4})^{2}}=\frac{x^{4} }{(1+x^{4})^{3}}+... =x^{4}*\frac{1+x^{4}}{x^{4}}=1+x^{4} \\\\x=3\\\\1+x^{4}=1+3^{4}=1+81=82\\\\Otvet:\boxed{82}](https://tex.z-dn.net/?f=x%5E%7B4%7D%2B%5Cfrac%7Bx%5E%7B4%7D%20%7D%7B1%2Bx%5E%7B4%7D%7D%2B%5Cfrac%7Bx%5E%7B4%7D%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B2%7D%7D%3D%5Cfrac%7Bx%5E%7B4%7D%20%7D%7B%281%2Bx%5E%7B4%7D%29%5E%7B3%7D%7D%2B...%20%3Dx%5E%7B4%7D%2A%5Cfrac%7B1%2Bx%5E%7B4%7D%7D%7Bx%5E%7B4%7D%7D%3D1%2Bx%5E%7B4%7D%20%5C%5C%5C%5Cx%3D3%5C%5C%5C%5C1%2Bx%5E%7B4%7D%3D1%2B3%5E%7B4%7D%3D1%2B81%3D82%5C%5C%5C%5COtvet%3A%5Cboxed%7B82%7D)
2)
![y_{n}=\frac{13-n}{5n+8}\\\\y_{n}=\frac{5}{48}\\\\\frac{13-n}{5n+8}=\frac{5}{48}\\\\48(13-n)=5(5n+8)\\\\624-48n=25n+40\\\\-48n-25n=40-624\\\\-73n=-584\\\\n=8\\\\Otvet:\boxed{n=8}](https://tex.z-dn.net/?f=y_%7Bn%7D%3D%5Cfrac%7B13-n%7D%7B5n%2B8%7D%5C%5C%5C%5Cy_%7Bn%7D%3D%5Cfrac%7B5%7D%7B48%7D%5C%5C%5C%5C%5Cfrac%7B13-n%7D%7B5n%2B8%7D%3D%5Cfrac%7B5%7D%7B48%7D%5C%5C%5C%5C48%2813-n%29%3D5%285n%2B8%29%5C%5C%5C%5C624-48n%3D25n%2B40%5C%5C%5C%5C-48n-25n%3D40-624%5C%5C%5C%5C-73n%3D-584%5C%5C%5C%5Cn%3D8%5C%5C%5C%5COtvet%3A%5Cboxed%7Bn%3D8%7D)
Подставляем
5 * 2 - 4 = 10 - 4 = 6
Правильный ответ 2.
Любое число в квадрате будет положительное, значит:
х²>64
x∈(-∞;-8)(8;+∞)
Предположим х=-9, тогда х²=(-9)²=81, значит х²>64 и т.д.
Отсюда получаем, что неравенство верное при х∈(-∞;-8)(8;+∞)
√27*tg(pi+3pi/4)*sin(8pi+2pi/3) = 3√3*tg(3pi/4)*sin(2pi/3) =
= 3√3*(-1)*√3/2 = -3*3/2 = -9/2 = -4,5