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

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

因为:a/(ab+a+1)=a/(ab+a+abc)=1/(b+1+bc)
可知: a/(ab+a+1)=1/b *b/(bc+b+1)
又有:c/(ca+c+1)=c/(ca+c+abc)=1/(a+1+ab)=1/a* a/(ab+a+1)
=1/a* 1/b *b/(bc+b+1)=1/ab* b/(bc+b+1)
所以:a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)
=1/b *b/(bc+b+1)+b/(bc+b+1)+1/ab* b/(bc+b+1)
=(a+ab+abc)/(a+ab+abc)
=1

因为:a/(ab+a+1)=a/(ab+a+abc)=1/(1+b+bc)
c/(ab+c+1)=bc/(abc+bc+b)=bc/(1+b+bc)
所以:a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)
=1/(1+b+bc)+b/(1+b+bc)+bc/(1+b+bc)
=(1+b+bc)/(1+b+bc)=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+...

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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

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