Question

(2的平方+1)(2的4次方+1)(2的8次方+1)(2的16次方+1)(2的32次方+1)=？

要简便运算。

Answer 1

解析

原式化简方法是乘以一个2^2-1，再除以一个2^2-1，利用平方差公式化简
(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)/(2^2-1)
=[(2^4-1)(2^4+1)](2^8+1)(2^16+1)(2^32+1)/(2^2-1)
=[(2^8-1)(2^8+1)](2^16+1)(2^32+1)/(2^2-1)
=[(2^16-1)(2^16+1)](2^32+1)/(2^2-1)
=[(2^32-1)(2^32+1)]/(2^2-1)
=(2^64-1)/(2^2-1)
=1/3(2^64-1)

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Answer 2

(2的平方+1)(2的4次方+1)(2的8次方+1)(2的16次方+1)(2的32次方+1)
=（2²-1）(2的平方+1)(2的4次方+1)(2的8次方+1)(2的16次方+1)(2的32次方+1)/3
=(2的64次方-1）/3

Answer 3

(2的平方+1)(2的4次方+1)(2的8次方+1)(2的16次方+1)(2的32次方+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)/(2^2-1)
=(2^4-1))(2^4+1)(2^8+1)(2^16+1)(2^32+1)/(2^2-1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)/(2^2-1)
=(2^16-1)(2^16+1)(2^32+1)/(2^2-1)
=(2^32-1)(2^32+1)/(2^2-1)
=(2^64-1)/(2^2-1)
=(2^64-1)/3