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9/7Ⅹ2/3+6/7x2/3一2/3简便计算

9/7Ⅹ2/3+6/7x2/3一2/3 =(9/7+6/7-1)x2/3 =8/7x2/3 =16/21

2.9*6.7+2.9+2.9*2.3 =2.9*(6.7+1+2.3) =2.9*10 =29

8/9+6/7×2/3+3/7 =8/9+(4/7+3/7) =8/9+1 =1又8/9。

=(1-1/2)+(1-1/3)+...(1-1/10) =9-(1/2+1/3+...1/10) =9-[1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10] =9-[(1/2+1/3+1/6)+(1/4+1/5+1/10)+1/7+1/8+1/9] =9-(1+11/20+1/8+1/7+1/9) =9-(1+27/40+1/9+1/7) =9-(1+283/360+1/7) =8-2341/2520

写得太复杂了,这样既可。int main(){ int N,i,sign=-1; float sum = 0; scanf("%d",&N); for(i = 1;i

首位相加: 1+100,2+99+……50+51 最后是101*50=5050。 当然如果学过了高斯求和,直接代公式就可以了: 高斯求和公式是:1+2+3+4+…+n=n(n+1)/2; 答案是一样的。 扩展资料: 文字表述:和=(首项 + 末项)x项数 /2数学表达:1+2+3+4+……+ n = (n+1)n ...

(1/7-1/9)乘3/6-2/3 =(9/63-7/63)x1/2-2/3 =2/63x1/2-2/3 =1/63-42/63 =-41/63

变形:1²/(2×3),2²/(3×4),3²/(4×5),4²/(5×6),5²/(6×7),……规律:从第1项开始,每一项的分子为项数的平方,分母为项数+1与项数+2的乘积。原数列为各分数化简为最简分数的结果。后面一项=6²/(7×8)=36/5

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