设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+2y=a,x,y∈R},若A∩B=φ,求a的值.
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设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+2y=a,x,y∈R},若A∩B=φ,求a的值.设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+
设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+2y=a,x,y∈R},若A∩B=φ,求a的值.
设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+2y=a,x,y∈R},若A∩B=φ,求a的值.
设A={(x,y)丨2x+y=1,x,y∈R},B={(x,y)丨a^2x+2y=a,x,y∈R},若A∩B=φ,求a的值.
A∩B=φ即两直线没有交点
2x+y=1
y=1-2x
a²x+2y=a
则a²x+2-4x=a
(a²-4)x=a-2
没有交点则这个方程应该无解
所以x系数为0而常数不等于0
所以a=-2
A和B都表示二条直线上的点,如A∩B=φ,说明二直线平行.
所以有:a^2/2=2/1不=a/1
解得:a=-2.
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