2+2+二甲基吡啶胺
茹逸虹40111/2+1/3+2/3+1/4+2/4+3/4+…+1/60+2/60+…+59/60 -
凌毓钧13140477374 ______ 1/2+1/3+2/3+1/4+2/4+3/4+…+1/60+2/60+…+59/60 =1/2+2/2+3/2+……+59/2 =(1/2+59/2)*59/2 =30*59/2 =885
茹逸虹4011求 lim n→∞ 2 n 2 +n+7 5 n 2 +4 . -
凌毓钧13140477374 ______ 答: lim n→∞ 2 n 2 +n+7 5 n 2 +4 = lim n→∞ 2+ 1 n + 7 n 2 5+ 4 n 2 = 2 5 .
茹逸虹4011简便算法(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/6) - (1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)怎么算 -
凌毓钧13140477374 ______ 原式=(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/6)-(1+1/2+1/3+1/4)*(1/2+1/3+1/4)-(1/5)*(1/2+1/3+1/4) =(1+1/2+1/3+1/4)*(1/6)-(1/5)*(1/2+1/3+1/4) =1/6+(1/2+1/3+1/4)*(1/6)-(1/5)*(1/2+1/3+1/4) =1/6+(1/2+1/3+1/4)*(-1/30) =47/360
茹逸虹40111/1*2+2/1*2*3+3/1*2*3*4+...+9/1*2*3*4*5*6*7*8*9*10 -
凌毓钧13140477374 ______ 1/1*2+2/1*2*3+3/1*2*3*4+4/1*2*3*4*5+...+9/1*2*3*4*5*6*7*8*9*10 =(2-1)/1*2+(3-1)/1*2*3 +(4-1)/1*2*3*4 +(5-1)/1*2*3*4*5 +...+(10-1)/1*2*3*4*5*6*7*8*9*10 =2/1*2-1/1*2+3/1*2*3-1/1*2*3 +...+10/1*2*3*4*5*6*7*8*9*10-1/1*2*3*4*5*6*7*8*9*10 =1/1-1...
茹逸虹40111*2*3+2*4*6+7*14*21/1*3*5+2*6*10+7*21*35 -
凌毓钧13140477374 ______ (1*2*3+2*4*6+7*14*21)/1*3*5+2*6*10+7*21*35)=((1*2*3+2*(1*2*3)+7*(1*2*3))/((1*3*5+2*(1*3*5)+7*(1*3*5))=((1*2*3)*(1+2+7))/((1*3*5)*(1+2+7))=2/5
茹逸虹4011计算(1+1/2+1/3+1/4)*(1+1/2+1/3+1/4+1/5) - (1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4) -
凌毓钧13140477374 ______ (1+1/2+1/3+1/4)*(1+1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)=(1+1/2+1/3+1/4+1/5)*(1+1/2+1/3+1/4-(1/2+1/3+1/4)=1+1/2+1/3+1/4+1/5=137/60
茹逸虹40111+1/(1+2)+1/(1+2+3)+.......+1/(1+2+3+.......100)= -
凌毓钧13140477374 ______ 解: 注意常用的关系式: 1/{n(n+1)/2}=2{1/n-1/(n+1)} 可以得到: 1/3=1/2-1/6, 1/(1+2+3)=1/6, 1/(1+2+3+4)=1/6-1/15, .... 所以 原式= 1+1/2-1/6+1/6+1/6-1/15+1/15_+1/12-...... =1+1/2+1/6+1/12+1/20+...+1/ 2450+1/(1+2+3+...+100) =1+1/2+1/2-1/3+...
茹逸虹4011我国某些地区喜食的腌制食品含有二甲基亚硝胺,其化学式为 (CH3)2NNO,它与导致癌症高发有关,它由 - --- -
凌毓钧13140477374 ______ 根据二甲基亚硝胺的化学式 (CH3)2NNO,分析可知其由C,H,N,O 4种元素组成; 根据标在元素符号右下角的数字表示一个分子中所含原子的个数;可知组成元素的原子个数比为: C:H:N:O=2:6:2:1; 根据相对分子的质量为组成分子的各原子的相对原子质量之和,可得二甲基亚硝胺相对分子质量为: 15*2+14*2+16=74; 故答案为:4; C:H:N:O=2:6:2:1;74.
茹逸虹4011(1*2*3+2*4*6+4*8*12+7*14*21)÷(1*3*5+2*6*10+4*12*20+7*21*35) -
凌毓钧13140477374 ______ 1*3*5+2*6*10+7*21*35分之1*2*3+2*4*6+7*14*21 =[1*2*3*(1+2*2*2+7*7*7)/[1*3*5*(1+2*2*2+7*7*7)] =1*2*3/1*3*5 =2/5