$tìm MIN$ `A` = `|2`x` – `\frac{1}{3}` |` – `1` $và$ `\frac{3}{4}` `B` = `\frac{1}{3}` |`x` – `2`| + `2`|`3` – `\frac{1}{2}` . `y`| +`4`

0 bình luận về “$tìm MIN$ `A` = `|2`x` – `\frac{1}{3}` |` – `1` $và$ `\frac{3}{4}` `B` = `\frac{1}{3}` |`x` – `2`| + `2`|`3` – `\frac{1}{2}` . `y`| +`4`”

1. $A=|2x-\dfrac{1}{3}|-1\dfrac{3}{4}$

   $=|2x-\dfrac{1}{3}|-\dfrac{7}{4}$

   $|2x-\dfrac{1}{3}|\ge 0$

   $\Leftrightarrow A\ge \dfrac{-7}{4}$

   $\min A=\dfrac{-7}{4}\Leftrightarrow 2x-\dfrac{1}{3}=0$

   $\Leftrightarrow x=\dfrac{1}{6}$

   $B=\dfrac{1}{3}|x-2|+2|3-\dfrac{y}{2}|+4$

   $\dfrac{1}{3}|x-2|\ge 0$

   $2|3-\dfrac{y}{2}|\ge 0$

   $\Leftrightarrow \dfrac{1}{3}|x-2|+2|3-\dfrac{y}{2}|\ge 0$

   $\Leftrightarrow B\ge 4$

   $\min B=4\Leftrightarrow x-2=3-\dfrac{1}{2}y=0$

   $+) x-2=0\Leftrightarrow x=2$

   $+) 3-\dfrac{y}{2}=0\Leftrightarrow \dfrac{y}{2}=3\Leftrightarrow y=6$

   $\Rightarrow (x;y)=(2;6)$ thì $B\min$

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2. Đáp án:

   `Ở dưới`

    

   Giải thích các bước giải:

   `A` = | $2x$ – `\frac{1}{3}` | – `\frac{7}{4}`  

   ⇔ | $2x$ – `\frac{1}{3}` | ≥ 0

   ⇔ | $2x$ – `\frac{1}{3}` | – `\frac{7}{4}` ≥ `\frac{-7}{4}`

   ⇔ `A` ≥ `\frac{-7}{4}`

   `Dấu “=” xảy ra:`

   ⇔ $2x$ – `\frac{1}{3}` = $0$

   ⇔$2x$ = `\frac{1}{3}`

   ⇔ `x` = `\frac{1}{6}`

   `Vậy MIN A` = `\frac{-7}{4}` ⇔ `x` = `\frac{1}{6}`

   `B` = `\frac{1}{3}` | `x` – $2$ | + 2| 3 – `\frac{y}{2}` + $4$

   ⇔ `\frac{1}{3}` | `x` – $2$ | ≥ $0$

   ⇔ $2$ | $3$ – `\frac{y}{2}` | ≥ $0$

   ⇔ `\frac{1}{3}` | `x` – $2$ | + $2$ | 3 – `\frac{y}{2}` | ≥ $0$

   ⇔`\frac{1}{3}` | `x` – $2$ | + $2$ | 3- `\frac{y}{2}` | + $4$ ≥ $4$

   ⇔ `B` ≥ $4$

   $Dấu “=” xảy ra:$

   ⇒ `x` – $2$ = $0$

        3 – $\frac{y}{2}$ = $0$

   ⇒ `x` = $2$

        `y` = $6$

   `Vậy MIN B` = $4$ ⇔ `x`,`y` ∈ {$2$ ; $6$}

         

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