M={(x,y)|(y-3)/(x-2)=a+1}
(y-3)/(x-2)=a+1
即y-3=(a+1)(x-2)且x≠2
M表示直線y-3=(a+1)(x-2) 除去(2,3)點的集合
N={(x,y)|(a^2-1)x+(a-1)y=15}中,
表達(dá)式:(a^2-1)x+(a-1)y=15
a=1時,N=Φ,M∩Φ=Φ符合題意
a≠1時,即y=-(a+1)x+15/(a-1)
若M∩N=φ,有2種情況
1)直線平行,
則a+1=-(a+1),a=-1
兩直線分別為y=3和y=-15/2
2)直線y=-(a+1)x+15/(a-1)過(2,3)點
此時,-2(a+1)+15/(a-1)=3
即-2a²+2+15=3a-3
2a²+3a-20=0
解得a=-4或a=5/2
綜上,滿足條件的a值有4個為1,-1,-4,5/2
已知M={(x,y)|y-3/x-2=a+1},N={(x,y)|(a^2-1)x+(a-1)y=15,若M∩N=空集,則a等于多少,求詳解
已知M={(x,y)|y-3/x-2=a+1},N={(x,y)|(a^2-1)x+(a-1)y=15,若M∩N=空集,則a等于多少,求詳解
數(shù)學(xué)人氣:708 ℃時間:2020-05-09 11:26:27
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