求(1/8+1/24+1/48+1/80+1/120+1/168+1/224+1/288)*128,请问可以套用哪种列项
2个回答
[1/(2x4)+1/(4x6)+1/(6x8)+……+1/(16x18)]x128
=1/4x(1-1/2+1/2-1/3+……+1/8-1/9)x128
=1/4x8/9x128
=256/9
相关问题
(1/8+1/24+1/48+1/80+1/20+1/168+1/224+1/288)x128=?
(8/1+24/1+48/1+80/1+120/1+168/1+224/1)X16
(1/8+1/24+1/48+1/80+1/120+1/168)*56
42x(8/1+24/1+48/1+80/1+120/1+168/1)=?
(1/8+1/24+1/48+1/80+1/120+1/168)简便计算
42×([1/8]+[1/24]+[1/48]+[1/80]+[1/120]+[1/168])
1+3又1/8+5又1/24+7又1/48+9又1/80+11又1/120+13又1/168+15又1/224+17又1
121 *3/4-3/8*40-0.75 (1/8+1/24+1/48+1/80+1/120+1/168)*56
计算11/8+1/24+1/48+1/80+1/120
42*(8分之1+24分之1+48分之1+80分之1+120分之1+168分之1)