Tìm Min,max: A=x^4+1/(x^2+1)^2

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Tìm Min,max:
A=x^4+1/(x^2+1)^2

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Latifah 4 years 2020-11-01T03:31:37+00:00 1 Answers 127 views 0

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  1. Philomena
    0
    2020-11-01T03:33:13+00:00
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    *Min:

    `A=(x^4+1)/((x^2+1)^2)=(2x^4+2)/(2(x^2+1)^2)=(x^4−2x^2+1+x^4+2x^2+1)/(2(x^2+1)2)`

    `=((x^2−1)^2+(x^2+1)^2)/(2(x^2+1)^2)=((x^2−1)^2)/(2(x^2+1)^2)+1/2≥1/2`

     Dấu ‘=’ xảy ra `⇔x=±1`

    Vậy `Min=1/2` khi `x=+-1`

    *Max:

    `A=(x^4+1)/((x^2+1)^2)=(x^4+2x^2+1−2x^2)/((x^2+1)^2)=((x^2+1)^2−2x^2)/((x^2+1)^2)`

    `=1−(2x^2)/((x^2+1)^2)≤1`

    Dấu ‘=’ xảy ra `⇔x=0`

    Vậy `Max=1 khi x=0`

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )