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