meili 发表于 2022-10-27 15:23:49

(1)1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)(2)abc=1,计算1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)

<p>问题:(1)1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)(2)abc=1,计算1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)
<p>答案:↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">韩希昌的回答:<div class="content-b">网友采纳  (1)1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)  =2/1*2+2/2*3+2/3*4+.+2/100*101  =2(1/1*2+1/2*3+1/3*4+.1/100*101)  =2(1/1-1/2+1/2-1/3+1/3-1/4+.+1/100-1/101)  =2*(1-1/101)  =200/101  (2)abc=1,计算  1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)  =abc/(abc+a+ab)+abc/(1+b+bc)+1/(1+c+ca)  =bc/(bc+1+b)+abc/(1+b+bc)+1/(1+c+ca)  =(bc+abc)/(bc+1+b)+1/(1+c+ac)  =b(c+ac)/(bc+abc+b)+1/(1+c+ac)  =(c+ac)/(c+ac+1)+1/(1+c+ac)  =(c+ac+1)/(c+ac+1)  =1
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查看完整版本: (1)1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)(2)abc=1,计算1/(1+a+ab)+1/(1+b+bc)+1/(1+c+ca)