若x,y满足y

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若x,y满足y
若x,y满足y

若x,y满足y
∵被开方数要大于等于0
∴x+1≥0 且 -(x+1)²≥0
∴x+1=0 ,x=-1
∴y﹤0+0+5 ,y﹤5
原式=√(y-5)²+|x+1|
=|y-5|+|x+1|
=-(y-5)+0
=5-y

∵y＜√(x+1)-√(x+1)²+5
x+1＞0
∴y＜5
∴√(y-5)²+|x+1|=x+y-4

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