若方程(2m2+m-3)x+(m2-m)y-4m+1=0表示一条直线,则实数m满足( )A.m≠1 �

若方程(2m²+m-3)x+(m²-m)y-4m+1=0表示一条直线,则实数m满足( )

A.m≠1 B.m≠- C.m≠0 D.m≠1且m≠-且m≠0

答案

解析：由2m²+m-3=0得m=-或m=1,

由m²-m=0得m=0或m=1,

所以使所给方程表示一条直线的实数m应满足m≠1.故选A.

答案：A

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