√(a²-3a-b)+b²+2b+1=0∵√a²-3a-b≥0b²+2b+1=(b+1)²≥0∴a²-3a-b=0b²+2b+1=(b+1)²=0解得b=-1a²-3a=b=-1a²-3a-|b|=b-|b|==-1-1=-2