Question

(1+2)(1+2的平方)(1+2的4次方)(1+2的8次方)=？

Answer 1

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

提改尘示：前面加2-1，然后反复用肢空平方差公式核饥禅。

Answer 2

只需拦答要增加一项（者李2-1）首衡迟：
(1+2)(1+2^2)(1+2^4)(1+2^8)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1.

Answer 3

(1+2)(1+2^2)(1+2^4)(1+2^8)
=(1-2)(1+2)(1+2^2)(1+2^4)(1+2^8)/(1-2)
=(1-2^2)(1+2^2)(1+2^4)(1+2^8)/(1-2)
=(1-2^4)(1+2^4)(1+2^8)/茄陪(1-2)
=(1-2^8)(1+2^8)/(1-2)
=(1-2^16)/镇纳袭御兄(1-2)
=2^16-1

Answer 4

(1+2)(1+2^2)(1+2^4)(1+2^8)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1
配成平方亮早枣局差公敬岩雀式

Answer 5

现在前面×（1-2）即可 即配成平方差公式