因为a+b+c=2吧
于是(c-1)+(a-1)+(b-1)=-1
而且(a-1)(b-1)=ab-a-b+1
所以[(c-1)+(a-1)+(b-1)]/[(a-1)(b-1)(c-1)]=-1/[(ab-a-b+1)(c-1)]如果A+B+C不等于2呢？假设a+b+c=m则(c-1)+(a-1)+(b-1)=m-3而且(a-1)(b-1)=ab-a-b+1所以[(c-1)+(a-1)+(b-1)]/[(a-1)(b-1)(c-1)]=(m-3)/[(ab-a-b+1)(c-1)]