作业帮(（1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/04/27 21:39:35
(（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))
a^4+1/4=(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
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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
无意中看到你的提问 就去百度了下 一不小心儿 还给找到了 就复制来了 具体对不对 你自己个儿 验证去吧!
(（1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)
求[(1^4+1/4)(3^4+1/4)(5^4+1/4)……(19^4+1/4)]/[(2^4+1/4)(4^4+1/
（1^4/4)(3^4+1/4)...(19^4+1\4) \ (2^4+1/4)(2^4+1/4)(4^4+1/4).
(1^4+1/4)*(3^4+1/4)*.(19^4+1/4)/(2^4+1/4)*(4^4+1/4).(20^4+1/
计算：[1/4（1^4+3^4+5^4+……+19^4)]/[1/4(2^4+4^4+6^4+……+20^4)]
4 4 4 4 4=0 4 4 4 4 4=1 4 4 4 4 4=2 4 4 4 4 4=3 4 4 4 4 4=4
4 4 4 4=1 4 4 4 4=2 4 4 4 4=3 填标点符号
1/4,-3/4,-3/4,1/4,17/4,( )
1/3,2/3,1/3,1/4,2/4,3/4,2/4,1/4,( ),( ),( ),( ),( ),( ),( )
3 1/4 1/5 4/7
4-1
添运算符号和小括号4 4 4 4=1 4 4 4 4=4 1 2 3 4 5=1