Question

розвязати рівняння терміново

Answer 1

$S\displaystyle \bigg(\frac{1}{x(n+1)} +\frac{1}{x(n-1)}-\frac{1}{x}\bigg)=0\\ S=0;~~ x \in (-\infty;0)\cup(0;+\infty);~~~ n \in (-\infty;-1)\cup(-1;1)\cup(1;+\infty) \\ \\if~~ S\ne 0~~ and~~ x\in (-\infty;0)\cup(0;+\infty)~~ than \\ \\ \frac{1}{x(n+1)} +\frac{1}{x(n-1)}-\frac{1}{x}=0\\ \\ \frac{n-1+n+1-(n-1)(n+1)}{x(n+1)(n-1)} =0\\ \\ 2n-n^2+1=0\\ n^2-2n-1=0\\ D=8\\ n_{1,2}=1\pm\sqrt{2}$

В качестве x можно взять любое число, кроме 0.

Answer 2

$\frac{S}{x} (\frac{1}{n+1} +\frac{1}{n-1} ) = \frac{S}{x} \\ 1)S = 0; x \in R/\{0\}; n \in R/\{-1; 1\}\\ 2) S\in R/\{0\}; x \in R/\{0\};\\ \frac{1}{n+1}+\frac{1}{n-1} =1\\ \frac{2n}{n^2-1} =1\\ n^2-1=2n\\ n^2-2n-1=0\\ D=4+4=8\\ n_1=\frac{2-2\sqrt{2}}{2} =1-\sqrt{2}\\ n_2=1+\sqrt{2}$