Question

如果a+b+c=0，a+1分之1+b+2分之1+c+3分之1=0，那么（a+1)2+(b+2)2+(c+3)2的值为?

要过程

Answer 1

∵A+B+C=0
∴(A+B+C)^2=0
∴-(A^2+B^2+C^2)=2(AB+BC+AC)
∵1/(A+1)+1/(B+2)+1/(C+3)=0
∴[(B+2)*(C+3)+(A+1)*(C+3)+(A+1)*(B+2)]/(A+1)(B+2)(C+3)=0
∵(A+1)(B+2)(C+3)≠0
∴(B+2)*(C+3)+(A+1)*(C+3)+(A+1)*(B+2)=0
∴2(AB+BC+AC)+10A+8B+
6C
+22=0
∴-(A^2+B^2+C^2)+10A+8B+6C+22=0
∴(A^2+B^2+C^2)=10A+8B+6C+22
(A+1)^2+(B+2)^2+(C+3)^2
=(A^2+B^2+C^2)+2A+4B+6C+14
=10A+8B+6C+22+2A+4B+6C+14
=12(A+B+C)+36
=12*0+36
=36
答
：(A+1)^2+(B+2)^2+(C+3)^2=36