已知|A+2B+3C|+(A-B+4 C)^2=0,ABC不等于0,则3A+2B-C/2A-3B+C=

问题描述:

已知|A+2B+3C|+(A-B+4 C)^2=0,ABC不等于0,则3A+2B-C/2A-3B+C=

|A+2B+3C|+(A-B+4C)^2=0
所以A+2B+3C=0 (1)
A-B+4C=0 (2)
(1)-(2)
3B-C=0
B=C/3
A=B-4C=-(11/3)C
所以原式=(-11C+2C/3-C)/(-22C/3-C+C)
=(-34C/3)/(-22C/3)
=17/11