初二数学计算题,急求(a+b+c)分之(a-b+c) -(b-c+a)分之（a-2b+3c）+（c-a-b)分之（b-2c）

原式=(a-b+c)/(a+b+c) - (a-2b+3c)/[-(c-a-b)] + (b-2c)/(c-a-b)=(a-b+c)/(a+b+c) + (a-2b+3c)/(c-a-b) + (b-2c)/(c-a-b)=(a-b+c)/(a+b+c) + (a-2b+3c+b-2c)/(c-a-b)=(a-b+c)/(a+b+c) + (a-b+c)/(c-a-b)=(a-b+c)(c-a...