Question

已知：abc=1,a+b+c=2,a的平方+b的平方+c的平方=3,求：1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)=?

Answer 1

(一）由题设条件得到的几个结论.（1）.因a+b+c=2,且a^2+b^2+c^2=3.===>4=(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)=3+2(ab+bc+ca).===>ab+bc+ca=1/2.(2)因a+b+c=2,===>ab+c-1=ab+1-a-b=(a-1)(b-1).同理有bc+a-1=(b-1)(c-1).ac+b-1=(a-1)(c-1).(3).(a-1)(b-1)(c-1)=abc+(a+b+c)-(ab+bc+ca)-1=1+2-(1/2)-1=3/2.(二）由结论（2）得,原式=[1/(a-1)(b-1)]+[1/(b-1)(c-1)]+[1/(c-1)(a-1)]=[(c-1)+(b-1)+(a-1)]/[(a-1)(b-1)(c-1)]=[(a+b+c)-3]/(3/2)=-2/3.