如果a+b+c>0,1/(a+1)+1/(b+2)+1/(c+3)=0,那么（a+1)^2+(b+2)^2+(c+3)^
1个回答
∵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
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