Question

已知abc=1，证明[1/ab+a+1]+[1/bc+b+1]+[1/ac+c+1]=1

人教版八年级下册数学分式运算

Answer 1

c=1/ab
原式=1/(ab+a+1)+1/(1/a+b+1)+1/(1/b+1/ab+1)
=1/(ab+a+1)+a/(1+ab+a)+ab/(a+1+ab)
=(ab+a+1)/(ab+a+1)
=1

Answer 2

ab+a+1=ab+a+abc=a(bc+b+1)

ac+c+1=ac+c+abc=c(ab+a+1)=ac(bc+b+1)

[1/ab+a+1]+[1/bc+b+1]+[1/ac+c+1]=1/(bc+b+1) * (1/a+1+1/ac)=1/(bc+b+1) * [(1+c+ac)/ac]=1

Answer 3

证明:
1/(ab+a+1)=c/(ab+a+1)c=c/abc+ac+c=c/(1+ac+c)
1/(bc+b+1)=a/(bc+b+1)a=a/abc+ab+a=a/(1+ab+a)=ac/(1+ac+c)
原式=c/(1+ac+c)+ac/(1+ac+c)+1/(1+ac+c)=(1+ac+c)/(1+ac+c)=1
证毕