若x*2+y*2-2x-6y+10=0,求（x*2-y*2）/xy=?

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若x*2+y*2-2x-6y+10=0,求（x*2-y*2）/xy=?
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若x*2+y*2-2x-6y+10=0,求（x*2-y*2）/xy=?
若x*2+y*2-2x-6y+10=0,求（x*2-y*2）/xy=?

若x*2+y*2-2x-6y+10=0,求（x*2-y*2）/xy=?
(x²-2x+1)+(y²-6y+9)=0
(x-1)²+(y-3)²=0
平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个都等于0
所以x-1=0,y-3=0
x=1,y=3
所以原式=(1-9)/3=-8/3

x*2+y*2-2x-6y+10=0
(x-1)^2+(y-3)^2=0
x=1 y=3
（x*2-y*2）/xy=-8/3

因为x*2+y*2-2x-6y+10=0,
x^2-2x+1+y^2-6y+9=0,
(x^2-2x+1)+(y^2-6y+9)=0,
(x-1)^2+(y-3)^2=0
所以x-1=0,y-3=0,
解得x=1,y=3,代人得
（x*2-y*2）/xy
=（1-9）/3
=-8/3

x^2+y^2-2x-6y+10=0,求（x^2-y^2）/xy=?
x^2-2x+1+y^2-6y+9=0
(x-1)^2+(y-3)^2=0
(x-1)^2=0,(y-3)^2=0
x=1,y=3
(x^2-y^2)/xy
=(1^2-3^2)/(1*3)
=-8/3

x^2-2x+1+(y^2-6y+9)=0
(x-1)^2+(y-3)^2=0
(x-1)^2>=0，(y-3)^2>=0两者和为0，必分别为0
x=1,y=3
(x^2-y^2)/xy=-8/3

（x*2-2x+1）+（y*2-6y+9）=0
（x-1）²+（y-3）²=0
∴ x-1=0或y-3=0
∴ X=1，Y=3
把 X=1，Y=3代入（x*2-y*2）/xy=（1²-3²）／（1×3）=-8／3

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