tìm min: `A=(x-2019)^2 (x+2020)^2`

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tìm min: `A=(x-2019)^2 (x+2020)^2`

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Lệ Thu 5 years 2021-05-18T06:01:43+00:00 2 Answers 36 views 0

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  1. Calantha
    0
    2021-05-18T06:03:05+00:00
    Reply

    Đáp án:

     

    Giải thích các bước giải

    `A=(x-2019)^2 (x+2020)^2`

    `A=[(x-2019)(x+2020)]^2>=0 ∀x`

    Dấu `=` xảy ra `<=>[(x-2019)(x+2020)]^2=0`

    `<=>(x-2019)(x+2020)=0`

    `<=>`\(\left[ \begin{array}{l}x-2019=0\\x+2020=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=2019\\x=-2020\end{array} \right.\) 

    Vậy $Min_{A}=0$ `<=>x∈{2019,-2020}`

  2. Doris
    0
    2021-05-18T06:03:09+00:00
    Reply

    `A=(x-2019)^2(x+2020)^2`

    `=[(x-2019)(x+2020)]^2`

    Do `[(x-2019)(x+2020)]^2≥0`

    `⇒A≥0(∀x)`

    Dấu “=” xảy ra khi `[(x-2019)(x+2020)]^2=0`

    `⇔(x-2019)(x+2020)=0`

    `⇔` \(\left[ \begin{array}{l}x-2019=0\\x+2020=0\end{array} \right.\) 

    `⇔` \(\left[ \begin{array}{l}x=2019\\x=-2020\end{array} \right.\) 

    Vậy GTNN của `A=0` khi `x=2019` hoặc `x=-2020`

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