2. Tìm GTNN của – |2x + 3| – 1/2|3 – y| + 3/5 Các bn giúp mk nhoa !!!

Question

2. Tìm GTNN của
– |2x + 3| – 1/2|3 – y| + 3/5

Các bn giúp mk nhoa !!!
2-tim-gtnn-cua-2-3-1-2-3-y-3-5-cac-bn-giup-mk-nhoa

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Acacia 5 years 2020-11-04T22:38:47+00:00 2 Answers 95 views 0

Answers ( )

  1. Doris
    0
    2020-11-04T22:39:59+00:00
    Reply

    Đáp án:

    \[{A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
    x =  – \dfrac{3}{2}\\
    y = 3
    \end{array} \right.\]

    Giải thích các bước giải:

     Ta có:

    \(\begin{array}{l}
    A =  – \left| {2x + 3} \right| – \dfrac{1}{2}\left| {3 – y} \right| + \dfrac{3}{5}\\
     = \dfrac{3}{5} – \left( {\left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 – y} \right|} \right)\\
    \left. \begin{array}{l}
    \left| {2x + 3} \right| \ge 0,\,\,\,\forall x\\
    \left| {3 – y} \right| \ge 0,\,\,\,\forall y
    \end{array} \right\} \Rightarrow \left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 – y} \right| \ge 0,\,\,\,\,\forall x,y\\
     \Rightarrow \dfrac{3}{5} – \left( {\left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 – y} \right|} \right) \le \dfrac{3}{5},\,\,\,\forall x,y\\
     \Rightarrow {A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
    \left| {2x + 3} \right| = 0\\
    \left| {3 – y} \right| = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x =  – \dfrac{3}{2}\\
    y = 3
    \end{array} \right.
    \end{array}\)

    Vậy \({A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
    x =  – \dfrac{3}{2}\\
    y = 3
    \end{array} \right.\)

  2. minhkhoi
    0
    2020-11-04T22:40:25+00:00
    Reply

    Có:

    `-|2x+3| ≤ 0`

    `-1/2|3-y|≤0` (do `|3-y|≥0`)

    `=> – |2x + 3 | -1/2 |3-y| ≤ 0`

    `=> – |2x + 3 | -1/2 |3-y| +3/5≤ 3/5`

    Dấu “=” xảy ra `⇔2x+3=3-y=0`

    `⇔ x = -3/2 ; y= 3`

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