若方程x^IaI-1+(a-2)y=3是二元一次方程,则a的取值范围是? 求过程
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人气:161 ℃ 时间:2020-02-04 10:34:14
解答
∵方程x^(IaI-1)+(a-2)y=3是二元一次方程
∴x的指数|a|-1=1且y的系数,a-2≠0
∴|a|=2,且a≠2
解得a=-2
∴则a的取值范围是a=-2
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