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求(x+1)(x²+1)(x^4+1)(x^8+1)...(x^64+1)的值
人气:317 ℃ 时间:2020-03-26 08:11:22
解答
原式=(x-1)(x+1)(x²+1)(x^4+1)(x^8+1)...(x^64+1)/(x-1)
=(x²-1)(x²+1)(x^4+1)(x^8+1)...(x^64+1)/(x-1)
反复用平方差
=(x^128-1)/(x-1)
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