Question

急,已知abc=1,a+b+c=2,a的平方加b的平方加c的平方=3,求1/ab+c-1+1/bc+a-1+1/ac+b-1

已知abc=1,a+b+c=2,a的平方加b的平方加c的平方=3,求1/ab+c-1+1/bc+a-1+1/ac+b-1

Answer 1

解：原式=1/ab +1/bc+ 1/ac +a +b +c -1 -1 -1
=c+a+b+a+b+c-1-1-1
=2+2-3
=1
希望能帮到你，有什么不懂的可以追问，祝你学习愉快！

Answer 2

(a+b+c)^2=a^2+b^2+c^2+2ac+2bc+2ab=4 和 a^2+b^2+c^2=3
2ac+2bc+2ab＝1
ac+bc+ab＝1/2
a+b+c=2
c=2-a-b
1/(ab+c-1)=1/(ab+2-a-b-1)=1/(ab+1-a-b)=1/(a-1)(b-1)
则1/（ab+c-1）+1/（bc+a-1）+1/（ac+b-1）＝1/(a-1)(b-1)＋1/(b-1)(c-1)+1/(a-1)(c-1)
=(c-1)/(a-1)(b-1)(c-1)＋(a-1)/(b-1)(c-1)(a-1)+(b-1)/(a-1)(c-1)(b-1)
=(a+b+c-1-1-1)/(a-1)(c-1)(b-1)
=-1/(abc-ac-bc-ab+a+b+c-1)
=-1/(1-1/2+2-1)
=-2/3

Answer 3

这个答案是 1.... 绝对是1 放心 谢谢采纳

Answer 4

ab=1/c,ac=1/b,bc=1/a,a^2+b^2+c^2=3。把这四个式子代进要求的那个式子里就行了！

Answer 5

1