Question

х^47+x^46+...+x+1/x^15+x^14+...+x+1
нужно сократить дробь.

Answer 1

$\frac{x^{47}+...+1}{x^{15}+...+1} = \frac{x^{32}(x^{15}+...+1)+x^{16}(x^{15}+...+1)+(x^{15}+...+1)}{x^{15}+...+1} = \\ = \frac{(x^{15}+...+1)(x^{32}+x^{16}+1)}{x^{15}+...+1} =\boxed{x^{32}+x^{16}+1}$