若根号x+y-3+绝对值xy+x+y-2=0,求x/y+y/x根号x+y-3+绝对值xy+x+y-2=0,求x/y+y/x
问题描述：
若根号x+y-3+绝对值xy+x+y-2=0,求x/y+y/x
根号x+y-3+绝对值xy+x+y-2=0,求x/y+y/x
答
根号下的数必须大于或等于0
绝对值出来的数也必须大于或等于0
两个相加也应该必须大于或等于0
所以x+y-3=0 xy+x+y-2=0
则（x+y）=3 , xy=-1 , (x+y)^2-2xy=x^2+y^2=11
x/y+y/x =[ (x/y)*xy+(y/x)*xy ] /xy = (x^2+y^2)/xy =11/（-1）=-11
答
有题意得
x+y-3>=0
xy+x+y-2>=0
因为根号x+y-3+绝对值xy+x+y-2=0
则x+y-3=0
xy+x+y-2=0
xy=-1
x+y=3
x^2+y^2=11
x/y+y/x =(x^2+y^2)/xy=-11