解：∵a+b=c∴b²+c²-a²=2b²+2ab=2b(a+b)=2bca²+c²-b²=2a²+2ab=2a(a+b)=2aca²+b²-c²=-2ab∴原式=2bc/2bc+2ac/2ac+2ab/(-2ab)=1+1-1=1
