Question

доведіть тотожність 60б срочно​

Answer 1

Ответ:

$\frac{3}{1-x^2}+\frac{3}{1+x^2}+\frac{6}{1+x^4}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=\frac{48}{1-x^{32}}$

Пошаговое объяснение:

$\frac{3}{1-x^2}+\frac{3}{1+x^2}+\frac{6}{1+x^4}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{3(1+x^2)}{(1-x^2)(1+x^2)}+\frac{3(1-x^2)}{(1-x^2)(1+x^2)}+\frac{6}{1+x^4}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{3+3x^2+3-3x^2)}{1-x^4}+\frac{6}{1+x^4}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{6}{1-x^4}+\frac{6}{1+x^4}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{6(1+x^4)}{(1-x^4)(1+x^4)}+\frac{6(1-x^4)}{(1-x^4)(1+x^4)}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{6+6x^4+6-6x^4}{1-x^8}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{12}{1-x^8}+\frac{12}{1+x^8}+\frac{24}{1+x^{16}}=$

$\frac{12(1+x^8)}{(1-x^8)(1+x^8)}+\frac{12(1-x^8)}{(1-x^8)(1+x^8)}+\frac{24}{1+x^{16}}=$

$\frac{12+12x^8+12-12-x^8}{1-x^{16}}+\frac{24}{1+x^{16}}=$

$\frac{24}{1-x^{16}}+\frac{24}{1+x^{16}}=$

$\frac{24(1+x^{16})}{(1-x^{16})(1+x^{16})}+\frac{24(1-x^{16})}{(1-x^{16})(1+x^{16})}=$

$\frac{24+24x^{16}+24-24x^{16}}{1-x^{32}}=\frac{48}{1-x^{32}}$