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y=(x²-3x+2)/ (x²+2x-3)的值域是什么
人气:377 ℃ 时间:2020-04-24 20:01:02
解答
y=(x²-3x+2)/ (x²+2x-3)
= (x-1)(x-2)/[(x+3)(x-1)]
= (x-2)/(x+3)
= (x+3-5)/(x+3)
= 1 - 5/(x+3)
-5/(x+3)≠0,1 - 5/(x+3)≠1
值域(-∞,1),(1,+∞)
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