[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/17 11:15:39
[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
a^4+1/4
=(a^2+1/2)^2-a^2
=(a^2+a+1/2)(a^2-a+1/2)
=[(a+1/2)^2+1/4][(a-1/2)^2+1/4]
根据这个式子,原式可化为:
[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
={[(2+1/2)^2+1/4][(2-1/2)^2+1/4]*[(3+1/2)^2+1/4][(3-1/2)^2+1/4]*...*[(10+1/2)^2+1/4][(10-1/2)^2+1/4]}/{[(1+1/2)^2+1/4][(1-1/2)^2+1/4]*...*[(9-1/2)^2+1/4][(9+1/2)^2+1/4]}
消去一样的式子:
比如:[(2-1/2)^2+1/4]=[(1+1/2)^2+1/4]
...
所以:
=[(10+1/2)^2+1/4]/[(1-1/2)^2+1/4]
=61
不知道中间的消去有没有错,电脑上看的晕了~
希望我的回答让你满意
孩子别懒,叫计算器来,真不知道它是干嘛吃的?
[2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)】/[1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)]
1,2,4,6,
((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))
求[(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)]求[(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、2、4、3、6、4、( )、(
计算:[1/4(1^4+3^4+5^4+……+19^4)]/[1/4(2^4+4^4+6^4+……+20^4)]
填上合适的运算符号和括号2 2 2 2 2 = 0 2 2 2 2 2 = 14/4+4-4=1 4/4+4/4=2 4 4 4 4=3 4 4 4 4=44 4 4 4=5 4 4 4 4=6 4 4 4 4=7 4 4 4 4=84 4 4 4=9 4 4 4 4=10
1+6+6+8+4+2+6+6+5+4+4+1+4+7+45
1,2,4,4,1,( )
4 4 4 4=1 4 4 4 4=2 4 4 4 4=3 填标点符号
数学(2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)/(1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)分数线前后是一个整体
1+2+4+6@$456
1/2+1/2+4+1/2+4+6+.+1/2+4+6+...+100
1+1/2+4+1/2+4+6+1/2+4+6+8+.+1/2+4+6+8...+20
已知1^4+2^4+3^4+4^4=354,求2^4+4^4+6^4+8^4=?怎么计算的?
计算 1/2*4+1/4*6+.1/2006*2008
1/2*4+1/4*6+...+1/2012*2014=?
1/2*4+1/4*6+.+1/2008*2010