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lilihurry

Òø³æ (СÓÐÃûÆø)

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½â·½³Ì×é ¢Ùx2-y2-5x+y+8=0,  ¢Ú2xy-x-5y+1=0
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lilihurry

Òø³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
2Â¥: Originally posted by feixiaolin at 2014-03-01 21:29:18
Parameters  x, y;
Function x^2-y^2-5*x+y+8=0;
Function 2*x*y-x-5*y+1=0;

x=3, y=2;
x=2, y=-1;

Çë¸ø³ö¾ßÌå²½Ö裬¸´Êý¸ùÒ²Çó£¬Ð»Ð»£¡
3Â¥2014-03-01 22:21:19
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
Parameters  x, y;
Function x^2-y^2-5*x+y+8=0;
Function 2*x*y-x-5*y+1=0;

x=3, y=2;
x=2, y=-1;
2Â¥2014-03-01 21:29:18
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

zsm28

Ìú³æ (³õÈëÎÄ̳)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

¡ï ¡ï ¡ï ¡ï ¡ï
¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
lilihurry(feixiaolin´ú·¢): ½ð±Ò+5 2014-03-02 10:25:06
x^2 - y^2 - 5*x + y + 8 = 0     (1)
2*x*y - x- 5*y + 1 = 0      (2)
(1):    x^2 - 5x + (5/2)^2 - (y^2 + y + (1/2)^2) + 2 = 0  
(x -5/2)^2 - (y - 1/2)^2 + 2 = 0
(2x - 5)^2 - (2y - 1)^2 + 8 = 0        (3)
(2):    y(2x - 5) = x - 1,
y = (x - 1)/(2x - 5)          (4)
2y - 1 = 3/(2x - 5)   
Áî   2x - 5 = u
2y - 1 = 3/u
´úÈë £¨3£©
u^2 - 9/u^2 + 8 = 0
u^4 + 8u^2 - 9 = 0
(u^2 - 1)(u^2 + 9) = 0
u = 1, -1, 3i, -3i
u = 2x - 5 = 1, x = 3, y = (3/u + 1)/2 = 2
u = 2x - 5 = -1, x = 2, y = (3/u + 1)/2 = -1
u = 2x - 5 = 3i, x = (5 + 3i)/2, y = (3/u + 1)/2 = (1 - i)/2
u = 2x - 5 = -3i, x = (5 - 3i)/2, y = (3/u + 1)/2 = (1 + i)/2
4Â¥2014-03-02 09:54:18
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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