更新1:
y^+5y+2x-xy+13=0的因式分解~~急10點
y^+5y+2x-xy+13=0的因式分解?
回答 (1)
2015-10-20 4:09 pm
y² + 5y + 2x - xy + 13 = 0
y² + 5y + 13 = x(y - 2)
x = (y² + 5y + 13) / (y - 2)
x = (y² + 5y -14 + 27) / (y - 2)
x = ( (y - 2)(y + 7) + 27 ) / (y - 2)
x = y+7 + 27/(y - 2)
若 x 是整數, 則 27/(y - 2) 是整數, y - 2 = ±1 , ±3 , ±9 , ±27
⇒ y = 1 , 3 , -1 , 5 , -7 , 11 , -25 , 29
⇒ (x,y) = (-19,1) , (37,3) , (-3,-1) , (21,5) , (-3,-7) , (21,11) , (-19,-25) , (37,29)
y² + 5y + 13 = x(y - 2)
x = (y² + 5y + 13) / (y - 2)
x = (y² + 5y -14 + 27) / (y - 2)
x = ( (y - 2)(y + 7) + 27 ) / (y - 2)
x = y+7 + 27/(y - 2)
若 x 是整數, 則 27/(y - 2) 是整數, y - 2 = ±1 , ±3 , ±9 , ±27
⇒ y = 1 , 3 , -1 , 5 , -7 , 11 , -25 , 29
⇒ (x,y) = (-19,1) , (37,3) , (-3,-1) , (21,5) , (-3,-7) , (21,11) , (-19,-25) , (37,29)
收錄日期: 2021-04-18 14:09:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151020063603AA4h2HZ