已知abc=1,求分式的通分,并求他们的和.

/>a/(ab+a+1)
=a/(ab+a+1)
b/(bc+b+1)
=ab/(abc+ab+a) (上下同时乘以a)
=ab/(ab+a+1)
c/(ac+c+1)
=bc/(abc+bc+b) （上下同时乘以ba）
=bc/(bc+b+1)
=1/(1+1/c+1/bc) 上下同除bc
=1/(ab+a+1)
∴a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=(a+ab+1)/(ab+a+1)
=1
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