先在前面乘以（2-1），然后连续运用平方差公式。
(2+1)(2^2+1)(2^4+1)(2^8+1)（2^16+1）－2^32
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)（2^16+1）－2^32
=(2²-1)(2²+1)(2^4+1)(2^8+1)（2^16+1）－2^32
=(2^4-1)(2^4+1)(2^8...
(2+1)(2^2+1)(2^4+1)(2^8+1)（2^16+1）－2^32
2个回答
相关问题
-
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1）
-
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
-
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
-
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
-
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/（2^32）-1=?
-
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(
-
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
-
求值:(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)-2^32
-
（2+1)(2²+1)(2^4+1)(2^8+1）（2^16+1）（2^32+1）的解
-
（1+2^(1/32))*(1+2^(1/16))*(1+2^(1/8))*(1+2^(1/4))*(1+2^(1/2)