已知60/(x+1)(x−2)(x+3)=A/x+1 +B/x−2+C/x+3，其中A，B，C为常数，则A+B+C=_．

∵Ax+1 +Bx−2+Cx+3=A(x−2)(x+3)+B(x+1)(x+3)+C(x+1)(x−2)(x+1)(x−2)(x+3)=A(x2+x−6)+B(x2+4x+3)+C(x2−x−2)(x+1)(x−2)(x+3)=(A+B+C)x2+(A+4B−C)x−6A+3B−2C(x+1)(x−2)(x+3)=60(x+1)(x−2)(x+3)，∴A+...