这个问题要怎么答？

無聊人

(x^2 + y^2 - 6)y + (xy + 5)2x = 0 --------- (1)
(x^2 + y^2 - 6)x + (xy + 5)2y = 0---------- (2)

我看到的解答是以graph的方式的

dunwan2tellu

(x^2 + y^2 - 6)y + (xy + 5)2x = 0 --------- (1)
(x^2 + y^2 - 6)x + (xy + 5)2y = 0---------- (2)

(1) * x - (2) * y : 2(xy+5)(x+y)(x-y) = 0 ......(3)

=> xy+5 = 0 OR x = y OR x = -y
case(i) xy+5 = 0
(1) : (x^2 + y^2 - 6)y = 0
=> x^2 + y^2 - 6 = 0 OR y = 0 .
when y = 0 , xy+5 = 5 =/= 0 (no solution)
so x^2 + y^2 = 6 , xy = - 5
=> (-5/y)^2 + y^2 = 6 ==> y^4 - 6y^2 + 25 = 0 =>(y^2-3)^2+16 > 0 No solution .

case(ii) : x = -y
(1): (y^2 + y^2 - 6)y - (-y^2 + 5)2y = 0
=> 4y(y+2)(y-2) = 0
when y = 0 => x = 0
when y = -2 , x = 2
when y = 2,x=-2

case(iii) : x = y
(1) :  (x^2 + x^2 - 6)x + (x^2 + 5)2x = 0
=> 4x(x^2+1) = 0 => x = 0

Answer : (x,y) = (0,0),(2,-2),(-2,2)

無聊人

原帖由 dunwan2tellu 于 9-9-2006 11:00 AM 发表

(1) * x - (2) * y : 2(xy+5)(x+y)(x-y) = 0 ......(3)

=> xy+5 = 0 OR x = y OR x = -y
case(i) xy+5 = 0
(1) : (x^2 + y^2 - 6)y = 0
=> x^2 + y^2 - 6 = 0 OR y = 0 .
when y = 0 , xy+5 = 5 ...

谢谢你的解答！