(1^4/4)(3^4+1/4)...(19^4+1\4) \ (2^4+1/4)(2^4+1/4)(4^4+1/4).
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/19 22:01:10
(1^4/4)(3^4+1/4)...(19^4+1\4) \ (2^4+1/4)(2^4+1/4)(4^4+1/4)...(20^4+1/4)=?
分子第一项:(1^4+1/4)
分子第一项:(1^4+1/4)
x^4+1/4=x^4+x^2+1/4-x^2=(x^2+1/2)^2-x^2)
=(x^2+1/2+x)(x^2+1/2-x)=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
即得x^4+1/4=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
(1^4+1/4)*(3^4+1/4)*.(19^4+1/4)/(2^4+1/4)*(4^4+1/4).(20^4+1/4)
={((1/2)^2+1/4)((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)...((37/2)^2+1/4)((39/2)^2+1/4)}/{((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)((9/2)^2+1/4)...((39/2)^2+1/4)((41/2)^2+1/4)}
=((1/2)^2+1/4)/((41/2)^2+1/4)=2/(41^2+1)=1/841
=(x^2+1/2+x)(x^2+1/2-x)=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
即得x^4+1/4=(x+1/2)^2+1/4)(x-1/2)^2+1/4)
(1^4+1/4)*(3^4+1/4)*.(19^4+1/4)/(2^4+1/4)*(4^4+1/4).(20^4+1/4)
={((1/2)^2+1/4)((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)...((37/2)^2+1/4)((39/2)^2+1/4)}/{((3/2)^2+1/4)((5/2)^2+1/4)((7/2)^2+1/4)((9/2)^2+1/4)...((39/2)^2+1/4)((41/2)^2+1/4)}
=((1/2)^2+1/4)/((41/2)^2+1/4)=2/(41^2+1)=1/841
((1^4+1/4)(3^4+1/4)(5^4+1/4).(19^4+1/4))/((2^4+1/4)(4^4+1/4)
(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/
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 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
计算:[1/4(1^4+3^4+5^4+……+19^4)]/[1/4(2^4+4^4+6^4+……+20^4)]
1/3,2/3,1/3,1/4,2/4,3/4,2/4,1/4,( ),( ),( ),( ),( ),( ),( )
1/4,-3/4,-3/4,1/4,17/4,( )
1/2,1/4,
“1”“2”“3”“4”
1+2+3×4