已知abc=1,求a/(ab+a+1)+b/(bc+b+1)=c/(ac+c+1)的值 .

熱心網(wǎng)友

因?yàn)閍bc=1,所以c=1/abb/(bc+b+1)=b/[(1/a+b+1)]=ab/(1+a+ab)c/(ac+c+1)=(1/ab)/[(1/b)+(1/ab)+1]=1/(1+a+ab)所以a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)　＝a/(ab+a+1)+ab/(1+a+ab)+1/(1+a+ab)　＝（ab+a+1)／（ab+a+1)　＝１

熱心網(wǎng)友

本題應(yīng)該是求a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)的值,如果是應(yīng)該這樣計(jì)算:原式=ac/(ab+a+1)*c+b*(a*c)/(bc+b+1)*(a*c)+c/(ac+c+1) =ac/(abc+ac+c)+abc/(abc*c+abc+ac)+c/(ac+c+1) =ac/(1+ac+c)+1/(c+1+ac)+c/(ac+c+1) (將abc=1代入) =(ac+1+c)/(ac+1+c) (分母相同,分子相加) =1

熱心網(wǎng)友

呵呵，寫(xiě)錯(cuò)了吧？再抄一遍上來(lái)吧！