Question

急!求教一道初中数学题!!!

如图,利用完全平方差公式解.

Answer 1

(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/3
=(2^8-1)(2^8+1)(2^16+1)/3
=(2^16-1)(2^16+1)/3
=(2^32-1)/3

Answer 2

原式=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/3
=(2^8-1)(2^8+1)(2^16+1)/3
=(2^16-1)(2^16+1)/3
=(2^32-1)/3

Answer 3

等式乘以2^2-1再除以2^2-1 2^2表示2的2次方

这样(2^2-1)(2^2+1)=2^4-1
同理(2^4-1)*(2^4-1)=2^8-1
(2^8-1)*(2^8+1)=2^16-1
(2^16-1)*(2^16+1)=2^32-1

最后除2^2-1就等于 (2^32-1)/(2^2-1)

Answer 4

解：原式=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/3
=(2^8-1)(2^8+1)(2^16+1)/3
=(2^16-1)(2^16+1)/3
=(2^32-1)/3

Answer 5

原式＝(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
再逐步利用利用完全平方差公式
＝(2^4-1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
=...
=(2^32-1)/(2^2-1)

Answer 6

已经这么多人回答了
我就不多说什么了~
(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/3
=(2^8-1)(2^8+1)(2^16+1)/3
=(2^16-1)(2^16+1)/3
=(2^32-1)/3
好好看看吧~！