若x-1的绝对值+[y-3]²=0,计算，[xy-X²]除以X²-2xy+y²/xy乘以x-y/X²
\u5df2\u77e5x+3\u5206\u4e4bx\u7684\u7edd\u5bf9\u503c-3\u7684\u503c\u4e3a0\uff0cy-1\u5206\u4e4by\u65e0\u610f\u4e49\uff0c\u6c42xy-y²\u5206\u4e4bx²+2xy+y{ |x|-3} / (x+3)=0\uff0c|x|-3=0\u4e14x+3\u22600\uff0c\u5f97x=3
y/(y-1)\u65e0\u610f\u4e49\uff0c\u2234y-1=0\uff0cy=1
(x²+2xy+y)/(xy-y²)=(x+y)^2 /{y(x-y)} = (3+1)^2 /{1*(3-1)} =16/2=8
(x-5)^2=8y-y^2-16
(x-5)^2=-(y-4)^2
(x-5)^2+(y-4)^2=0
\u7531\u4e8e\u5e73\u65b9\u9879\u6052\u975e\u8d1f\uff0c\u56e0\u6b64\u4e24\u5e73\u65b9\u9879\u5747=0
x-5=0 x=5
y-4=0 y=4
x=5 y=4
y=3
[xy-X²]/X²-2xy+y²/xy*x-y/X²
=2-6 +3-3=-4
|X-1|+（Y-3）²=0
X-1=0，X=1‘；Y-3=0，Y=3
（XY-X²）÷（X²-2X+Y²）/XY×（X-Y）/X²
=X（Y-X）×XY/（X-Y）²×（X-Y）/X²
=-Y
代入Y=3
原式=-3
解答：
|x-1|+(y-3)^2=0，所以：x-1=0且y-3=0，解得：x=1，y=3
(xy-x^2)除[(x^2-2xy+y^2)/xy]*(x-y)/x^2
=x(y-x)*xy/(x-y)^2*(x-y)/x^2
=-y=-3
由已知条件可知:X=1 Y=3
则那个复杂的算式结果是:2
绛旓細鍥炵瓟锛x-1=0,y-3=0,鎵浠=1,y=3,-x-y=-4
绛旓細x=1 y=3 [xy-X²]/X²-2xy+y²/xy*x-y/X²=2-6 +3-3=-4
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