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

显示全部楼层 · 2011-12-18 17:14:32

题目已经提示了呀：“利用平方差公式”平方差公式是“(a-b)*(a+b)=a^2-b^2”对吧？但是观察题目里的式子，显然少了(a-b)这一项（因为题目里都是加号的项，却唯独没有减号项），因此，我们便来人为地添上一个减号——分子分母同乘(2-1)：原式=(2-1)*[(2+1)*(2^2+1)*(2^4+1)*(2^8+1)+1]/(2-1)
=[(2-1)*(2+1)*(2^2+1)*(2^4+1)*(2^8+1)+(2-1)*1]/1
=(2^2-1)*(2^2+1)*(2^4+1)*(2^8+1)+1
=(2^4-1)*(2^4+1)*(2^8+1)+1
=(2^8-1)*(2^8+1)+1...

千问 · 2011-12-18 17:14:32

解：因为(2-1)=1所以可以给原式乘上(2-1)，原式的值不变原式=(2-1)(2+1)(22+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=(22-1)(22+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^...

千问 · 2011-12-18 17:14:32

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=﹙2－1﹚(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=﹙22－1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=﹙2^4－1)(2^4+1)(2^8+1)(2^16+1)(2^32...

千问 · 2011-12-18 17:14:32

=（2-1）(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=（2^2 -1）(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=（2^32 -1）（2^32+1）=2^64 -1...

千问 · 2011-12-18 17:14:32

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