某同学在计算3(4+1)(4的平方+1)时,把33写成4-1后,发现可以连续运用平方差公式计算:

2024-12-20 18:14:35
推荐回答(2个)
回答1:

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

回答2:

=2