1/(A-B)(A-C)+1/(B-C)(B-A)+1/(C-A)(C-B)
问题描述:
1/(A-B)(A-C)+1/(B-C)(B-A)+1/(C-A)(C-B)
答
原式=(B-C)/(A-C)(A-B)(B-C)-(A-C)/(A-C)(A-B)(B-C)+(A-B)/(A-C)(A-B)(B-C)
=B-C-A+C+A-B/(A-C)(A-B)(B-C)
=0