230问答网 > [(1^4+1⼀4)(3^4+1⼀4)(5^4+1⼀4)......(19^4+1⼀4)]⼀[(2^4+1⼀4)(4^4+1⼀4)(6^4+1⼀4)......(20^4+1⼀4)]

[(1^4+1⼀4)(3^4+1⼀4)(5^4+1⼀4)......(19^4+1⼀4)]⼀[(2^4+1⼀4)(4^4+1⼀4)(6^4+1⼀4)......(20^4+1⼀4)]

2024-12-12 14:13:34

推荐回答（1个）

回答1：

你好，过程如下
其中的每一项可以看做
a^4+1/4
然后化简
=(a²+1/2)²-a²=(a²+a+1/2)(a²-a+1/2)=[a(a+1)+1/2][a(a-1)+1/2]

所以
[(1^4+1/4)(3^4+1/4)(5^4+1/4)……(19^4+1/4)]/[(2^4+1/4)(4^4+1/4)(6^4+1/4)……(20^4+1/4)]
=[(1*2+1/2)*(1*0+1/2)]·[(3*4+1/2)*(3*2+1/2)]·…·[(19*20+1/2)*(19*18+1/2)]/·[(2*3+1/2)*(2*1+1/2)]·[(4*5+1/2)*(4*3+1/2)]·…·[(20*21+1/2)*(20*19+1/2)]=（1*0+1/2）/（20*21+1/2)=1/841