Question

已知abc=1,求ab+a+1分之a+bc+b+1分之

已知abc=1,求(ab+a+1分之a) +(bc+b+1分之b) +(ac+c+1分之c)

Answer 1

a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=a/(ab+a+abc)+b/(bc+b+1)+c/(ac+c+abc) 利用abc=1
=a/a(b+1+bc)+b/(bc+b+1)+c/c(a+1+ab)
=a/a(b+1+bc)+b/(bc+b+1)+c/c(a+abc+ab) 利用abc=1
=a/a(b+1+bc)+b/(bc+b+1)+c/ac(1+bc+b)
=ac/ac(b+1+bc)+abc/ac(bc+b+1)+c/ac(1+bc+b) 通分
=(ca+abc+c)/ac(1+bc+b)
=(ac+1+c)/(ac+abc*c+abc)
=(ac+1+c)/(ac+c+1)
=1

Answer 2

a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=a/(ab+a+abc)+b/(bc+b+1)+c/(ac+c+abc) 利用abc=1
=a/a(b+1+bc)+b/(bc+b+1)+c/c(a+1+ab)
=a/a(b+1+bc)+b/(bc+b+1)+c/c(a+abc+ab) 利用abc=1
=a/a(b+1+bc)+b/(bc+b+1)+c/ac(1+bc+b)
=ac/ac(b+1+bc)+abc/ac(bc+b+1)+c/ac(1+bc+b) 通分
=(ca+abc+c)/ac(1+bc+b)
=(ac+1+c)/(ac+abc*c+abc)
=(ac+1+c)/(ac+c+1)
=1