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

来源：百度知道 编辑：UC知道 时间：2024/05/29 04:33:44

abc=1
所以b=1/ac
ab=1/c
bc=1/a

所以原式=a/(1/c+a+1)+(1/ac)/(1/a+1/ac+1)+c/(ac+c+1)
=ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1)
=(ac+c+1)/(ac+c+1)
=1

abc=1
所以b=1/ac
ab=1/c
bc=1/a

所以原式=a/(1/c+a+1)+(1/ac)/(1/a+1/ac+1)+c/(ac+c+1)
=ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1)
=(ac+c+1)/(ac+c+1)
=1

a/(ab+a+1)
=a/(ab+a+abc)
=1/(bc+b+1)

a/(ab+a+1)
=(ac)/(abc+ac+c)
=(ac)/(ca+c+1)

进行类似变换可得

3[a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)]
=a/(ab+a+1)+1/(bc+b+1)+(ac)/(ca+c+1)+b/(bc+b+1)+1/(ca+c+1)+(ab)/(ab+a+1)+c/(ca+c+1)+1/(ab+a+1)+(bc)/(bc+b+1)
=(ab+a+1)/(ab+a+1)+(bc+b+1)/(bc+b+1)+(ca+c+1)/(ca+c+1)
=1+1+1
=3

a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)=1

1