简便运算:(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*...*(1+1\/99)*(1-1\/99...
先把括号里的约分得到:3\/2*1\/2*4\/3*2\/3...*97\/98*99\/98*100\/99*98\/99 之后会发现中间所有之积得1,最后剩个1\/2和100\/99相乘 结果就是,你知道的50\/99
简便运算:(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*...*(1+1\/99)*(1-1\/99...
(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*...*(1+1\/99)*(1-1\/99)*(1+1\/100)*(1-1\/100)= 3\/2 * 1\/2 *4\/3 *2\/3 *5\/4 *3\/4 *……* 100\/99 * 98\/99 *101\/100 *99\/100 =1\/2*(3\/2*2\/3)*(4\/3*3\/4)*(5\/4*4\/5)……*(100\/99*99\/100)*101\/...
(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*...*(1+1\/99)*(1-1\/99) 怎样简便计算...
(1+1\/3)×(1﹣1\/4)=1 ……(1+1\/98)×(1﹣1\/99)=1 最后就剩(1﹣1\/2)和(1+1\/99)相乘 所以原式=50\/99
(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*.……(1+1\/99)*(1-1\/99)的简便...
(1+1\/2)*(1-1\/2)*(1+1\/3)*(1-1\/3)*...*(1+1\/99)(1-1\/99)=(1+1\/2)(1+1\/3)...(1+1\/99)(1-1\/2)(1-1\/3)...(1-1\/99)=[3\/2*4\/3...100\/99]*[1\/2*2\/3...98\/99]=100\/2*1\/99 =50\/99
简便计算:(1+2\/1)×(1-2\/1)×...(1+99\/1)×(1-99\/1)=?
若是不太理解,则请在第四行多写几个自然找到规律。供参考,请笑纳。
...解法:(1+1\/2)(1-1\/2)(1+1\/3)(1-1\/3)……(1+1\/99)(1-1\/99)_百度知 ...
化成这样 3\/2 1\/2 4\/3 2\/3 5\/4 然后看有什么规律 基本上上下可以抵消 最后只剩下几个数 3 1 4 2 5 3 6 4 ... 100 98 1 100 100 --- --- ==> --- ==>--- 2 2 3 3 4 4 5 5...
...1-1\/2)()×(1+1\/3)×(1-1\/3)×(1+1\/4)×(1-1\/4)×...(1+1\/99...
(1+1\/2)×(1-1\/2)()×(1+1\/3)×(1-1\/3)×(1+1\/4)×(1-1\/4)×...(1+1\/99)×(1-1\/99)=3\/2×1\/2×4\/3×2\/3×5\/4×3\/4×。。。×99\/98×97\/98×100\/99×98\/99 =1\/2×100\/99 =50\/99
(1+1\/2)乘(1-1\/2)乘(1+1\/3)乘(1-1\/3)乘...乘(1+1\/99)乘(1-1\/99)=...
利用交换律,将“-”的为一组,“+”的为一组 (1+1\/2)×(1-1\/2)×(1+1\/3)×(1-1\/3)……(1+1\/99)×(1-1\/99)= [(1+1\/2)×(1+1\/3)×(1+1\/4)×……×(1+1\/99)]×[(1-1\/2)×(1-1\/3)×(1-1\/4)×……×(1-1\/99)]=[(3\/2)×(4\/3)×(5\/4)×…...
如何巧算(1+1\/2)×(1-1\/2)×(1+1\/3)×(1-1\/3)...(1+1\/99)×?
(1+1\/2)×(1-1\/2)×(1+1\/3)×(1-1\/3)...(1+1\/99)=【(1+1\/2)×(1+1\/3)×(1+1\/4)...(1+1\/99)】×【(1-1\/2)×(1-1\/3)×(1-1\/4)...(1-1\/98)】=【3\/2×4\/3×5\/4×...100\/99】×【1\/2×2\/3×3\/4×4\/5×...×97\/98】=...
(1+1\/2)x(1-1\/2)x(1+1\/3)x(1-1\/3)x……x(1+1\/99)x(1-1\/
对所求代数式变形,将对应的减法与加法交换位置 求出每个因式的值,消去互为倒数的项 结果=50\/99 过程如下图: