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求y=[(x+5)(x+2)]/(x+1)(x大于等于1)的值域
人气:382 ℃ 时间:2020-02-01 10:21:10
解答
y=[(x+5)(x+2)]/(x+1)
=(x²+7x+10)/(x+1)
=(x²+7x+6+4)/(x+1)
=(x+1)(x+6)/(x+1)+4/(x+1)
=x+6+4/(x+1)
=(x+1)+4/(x+1)+5
≥2√[(x+1)×4/(x+1)]+5=4+5=9
当x+1=4/(x+1)时取等号
解得x=1,另外一个x=-3不符合题意
所以值域y≥9
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