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当X>1时‘求Y=(2X平方-2X+1)/(X-1)的最小值!
人气:283 ℃ 时间:2020-05-12 16:39:16
解答
Y=(2x^2-2x+1)/(x-1)
=(2x(x-1)+1)/(x-1)
=2x+1/(x-1)
=2(x-1)+1/(x-1)+2

因为x>1因此x-1和1/(x-1)为正数

所以最小值为2根号(2(x-1)*1/(x-1))+2

=2根号2+2
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