1/ab+c-1+1/bc+a-1+1/ca+b-1
=(c+abc^2-abc+a+a^2abc-abc+b+ab^2c-abc)/abc 通分
=[(a+b+c)+abc(a+b+c)-3abc]/abc 合并,移项
=(2+1*2-3)/1
=1
1/ab+c-1+1/bc+a-1+1/ca+b-1
=c/abc+a/abc+b/abc+a-1+b-1+c-1
=c+a+b+a+b+c-3
=2(a+b+c)-3
=4-3
=1
abc=1
a+b+c=2
(a+b+c)/abc=1/2
1/ab+1/bc+1/ac=1/2
原式=1/ab+1/bc+1/ac+a+b+c-3
=1/2+2-3=-1/2
1/ab+1/bc+1/ca+a+b+c-3
=a+b+c/abc+a+b+c-3
=2+2-3
=1
1
通分即可,将条件带入......
原式=(a+b+c)/abc+a+b+c-3=1
1,你那个是1/ab+c-1还是1/(ab+c-1)呀?如果是前者就是1,太简单了这样,前面的已知条件都没用上,如是后者我还没算,比较麻烦,你应该写细些