(2x+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)

Question

(2x+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)

Алгебра

Гость

Ответ(ы) на вопрос:

Accepted Answer

[latex](2x+1)(2^1-1)(2^21)(2^4+1)*2^8+1)(2 ^{16}+1)(2 ^{32} +1)/(2^2-1)= [/latex][latex](2x+1)(2^4-1)(2^4+1)(2^8+1)(2 ^{16} +1)(2 ^{32} +1)/3=[/latex][latex](2x+1)(2^8-1)(2^8+1)(2 ^{16}+1)(2 ^{32} +1)/3=(2x+1)(2 ^{16}-1)(2 ^{16} +1 [/latex][latex])(2 ^{32}+1)/3=(2x-1)(2 ^{32} -1)(2 ^{32} +1)/3=(2x+1)(2 ^{64} -1)/3[/latex]