若a>b>0>c,a+b+c=0,且|b+1|=|c+1|,求-a平方+2b-c-(a-4c-b)的值|b+1|=|c+1|,则有：b+1=-c-1 得：b+c=-2又：a+b+c=0 则有：a=2所以：-a平方+2b-c-(a-4c-b)=-a^2+2b-c-a+4c+b=-a^2-a+3(b+c)=-4-2-6=-12