(9+1)(9的二次方+1)(9的四次方+1)=K分之9的八次方-1 則K=?

👨🏻‍🏫 學術台數學

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2015-09-14 1:26 pm

回答 (1)

用戶45850

2015-09-14 1:34 pm

(9-1)(9+1)(9^2+1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^2-1)(9^2+1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^4-1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^8-1)/8=(9^8 -1)/k ==>k=8

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