根据图有:a+b<0,c-b>0,a+c<0
故:√(a+b)²-∣c-b∣+∣a+c∣
=∣a+b∣-∣c-b∣+∣a+c∣
=-(a+b)-(c-b)-(a+c)
=-a-b-c+b-a-c
=-2a-2c
因为√(x+1)和√(x-y)²互为相反数
故:√(x+1)+√(x-y)²=0
故:x+1=0,x-y=0
故:x=y=-1
故:(xy)^2009=1
根据图有:a+b<0,c-b>0,a+c<0
故:√(a+b)²-∣c-b∣+∣a+c∣
=∣a+b∣-∣c-b∣+∣a+c∣
=-(a+b)-(c-b)-(a+c)
=-a-b-c+b-a-c
=-2a-2c
因为√(x+1)和√(x-y)²互为相反数
故:√(x+1)+√(x-y)²=0
故:x+1=0,x-y=0
故:x=y=-1
故:(xy)^2009=1