5.21 一又3/5×1/7+1/21＝几？

一又3/5×1/7+1/21＝几？  

一又3/5×1/7+1/21＝几？

一又3/5=8/5,所以
原式=（8/5）x（1/7）+1/21
=8/35+1/21
=24/105+5/105
=29/105

1/3*5+1/5*7+1/7*9……+1/19*21

1/3*5+1/5*7+1/7*9……+1/19*21
=2*(1/3-1/5+1/5-1/7+...+1/19-1/21)
=2*(1/3-1/21)
=4/7

-3/7减-1/5减-2/7加负一又1/5怎样算?

-3/7减-1/5减-2/7加负一又1/5
=-3/7+1/5+2/7-1又1/5
=(-3/7+2/7)+(1/5-1又1/5)
=-1/7-1
=-1又1/7

1/1*3+1/3*5+1/5*7+1/7*9+........+1/19*21

1/1 - 1/3 = 3/3 - 1/3 = 2/(1×3)
1/3 - 1/5 = 5/(3×5) - 3/(3×5) = 2/(3×5)
……
1/19 - 1/21 = 21/(19×21) - 19/(19×21) = 2/(19×21)
所以，上式的和等于：
=1/2 ×[(1/1 - 1/3) +(1/3-1/5) +……+(1/19 - 1/21)]
=1/2 ×(1 - 1/21)
=1/2 ×20/21
=10/21

一又3/4÷3/7-1/4乘于2有1/3。

1又3/4÷3/7-1/4×2又1/3
=1又3/4×7/3-1/4×7/3
=(1又3/4-1/4)×7/3
=3/2×7/3
=7/2
=3又1/2

[3/7+1/2*(-2/3)]*(-21/5)

[3/7+1/2*(-2/3)]*(-21/5)
=[3/7-1/3]*(-21/5)
=-3/7*21/5+1/3*21/5
=-9/5+7/5
=-2/5

1/1×3＋1/3×5＋1/5×7＋1/7×9＋1/9×11＋……＋1/19×21

1/1×3＋1/3×5＋1/5×7＋1/7×9＋1/9×11＋……＋1/19×21
=(1/2)(2/1×3＋2/3×5＋……＋1/19×21)
=(1/2)(1/1-1/3+1/3-1/5+……＋1/19-1/21)
=(1/2)(1/1-1/21)
=10/21

(1/4)^2÷(-1/2)^3+0÷2/5-[21*(3又1/7-7又1/3)*7/22÷3/1/2]^2

(1/4)^2÷(-1/2)^3+0÷2/5-[21*(3又1/7-7又1/3)*7/22÷3又1/2]^2
=1/16÷（-1/8）+0-[21*(22/7-22/3)*7/22÷7/2]^2
=-1/2-[（21÷7/2）*(1/7-1/3)*22*7/22]^2
=-1/2-[6*(1/7-1/3)*7]^2
=-1/2-[(6*7*1/7-6/7*1/3)]^2
=-1/2-（6-14）^2
=-1/2-（-8）^2
=-1/2-64
=-64又1/2

已知1/3-1/5=1/3*5,计算:3*5+1/5*7+1/7*9+……+1/19*21

1/3*5+1/5*7+1/7*9+……+1/19*21
=(1/2)(1/3-1/5+1/5-1/7+.....+1/19-1/21)
=(1/2)(1/3-1/21)
=(1/2)(6/21)
=1/7