Tìm `GTNN` của : `|x-2018|+|x-2019|+|x-2020|+|y-2021|` 18/07/2021 Bởi Athena Tìm `GTNN` của : `|x-2018|+|x-2019|+|x-2020|+|y-2021|`
Đáp án: GTNN của `|x – 2018| + |x – 2019| + |x – 2020| + |y – 2021|` là `2 ⇔ x = 2019; y = 2021` Giải thích các bước giải: Đặt `A = |x – 2018| + |x – 2019| + |x – 2020| + |y – 2021|` `= (|x – 2018| + |x – 2020|) + |x – 2019| + |y – 2021|` `= (|x – 2018| + |2020 – x|) + |x – 2019| + |y – 2021|` `≥ |x – 2018 + 2020 – x| + 0 + 0 = 2` `⇒ A ≥ 2` Dấu “=” xảy ra `⇔` $\left\{ \begin{array}{l}(x – 2018)(2020 – x) ≥ 0\\|x – 2019| = 0\\|y – 2021| = 0\end{array} \right.$ `⇔` $\left\{ \begin{array}{l}(x – 2018)(x – 2020) ≤ 0\\x – 2019 = 0\\y – 2021 = 0\end{array} \right.$ `⇔` $\left\{ \begin{array}{l}(x – 2018)(x – 2020) ≤ 0\\x = 2019\\y = 2021\end{array} \right.$ +) Với `(x – 2018)(x – 2020) ≤ 0` `⇔ x – 2018` và `x – 2020` trái dấu Mà `x – 2018 > x – 2020` `⇔` $\left\{ \begin{array}{l}x – 2018 ≥ 0\\x – 2020 ≤ 0\end{array} \right.$ `⇔` $\left\{ \begin{array}{l}x ≥ 2018\\x ≤ 2020\end{array} \right.$ `⇔ 2018 ≤ x ≤ 2020` Kết hợp với $\left\{ \begin{array}{l}x = 2019\\y = 2021\end{array} \right.$ nên ta có: `x = 2019 ; y = 2021` Kết luận: … Bình luận
`|x-2018|+|x-2019|+|x-2020|+|y-2021|` `=|x-2018|+|2020-x|+|x-2019|+|y-2021|` `≥|x-2018+2020-x|+|x-2019|+|y-2021|` `=|2|+0+0` `=2` Dấu `=` xảy ra khi \begin{cases} (x-2018)(2020-x)\ge0 \\ x-2019=0 \\ y-2021=0 \end{cases} `↔` \begin{cases} \left[ \begin{array}{l}\left \{ {{x-2018\ge0} \atop {2020-x\ge0}} \right.\\\left \{ {{x-2018\le0} \atop {2020-x\le0}} \right.\end{array} \right. \\ x=2019 \\ y=2021 \end{cases} `↔` \begin{cases} \left[ \begin{array}{l}\left \{ {{x\ge2018} \atop {x\le2020}} \right.\\\left \{ {{x\le2018} \atop {x\ge2020}} \right.\end{array} \right. \\ x=2019 \\ y=2021 \end{cases} `↔` \begin{cases} 2018\le x \le2020 \\ x=2019 \\ y=2021 \end{cases} `↔ (x;y)=(2019;2021)` Bình luận
Đáp án:
GTNN của `|x – 2018| + |x – 2019| + |x – 2020| + |y – 2021|` là `2 ⇔ x = 2019; y = 2021`
Giải thích các bước giải:
Đặt `A = |x – 2018| + |x – 2019| + |x – 2020| + |y – 2021|`
`= (|x – 2018| + |x – 2020|) + |x – 2019| + |y – 2021|`
`= (|x – 2018| + |2020 – x|) + |x – 2019| + |y – 2021|`
`≥ |x – 2018 + 2020 – x| + 0 + 0 = 2`
`⇒ A ≥ 2`
Dấu “=” xảy ra `⇔` $\left\{ \begin{array}{l}(x – 2018)(2020 – x) ≥ 0\\|x – 2019| = 0\\|y – 2021| = 0\end{array} \right.$
`⇔` $\left\{ \begin{array}{l}(x – 2018)(x – 2020) ≤ 0\\x – 2019 = 0\\y – 2021 = 0\end{array} \right.$
`⇔` $\left\{ \begin{array}{l}(x – 2018)(x – 2020) ≤ 0\\x = 2019\\y = 2021\end{array} \right.$
+) Với `(x – 2018)(x – 2020) ≤ 0`
`⇔ x – 2018` và `x – 2020` trái dấu
Mà `x – 2018 > x – 2020`
`⇔` $\left\{ \begin{array}{l}x – 2018 ≥ 0\\x – 2020 ≤ 0\end{array} \right.$
`⇔` $\left\{ \begin{array}{l}x ≥ 2018\\x ≤ 2020\end{array} \right.$
`⇔ 2018 ≤ x ≤ 2020`
Kết hợp với $\left\{ \begin{array}{l}x = 2019\\y = 2021\end{array} \right.$
nên ta có: `x = 2019 ; y = 2021`
Kết luận: …
`|x-2018|+|x-2019|+|x-2020|+|y-2021|`
`=|x-2018|+|2020-x|+|x-2019|+|y-2021|`
`≥|x-2018+2020-x|+|x-2019|+|y-2021|`
`=|2|+0+0`
`=2`
Dấu `=` xảy ra khi
\begin{cases} (x-2018)(2020-x)\ge0 \\ x-2019=0 \\ y-2021=0 \end{cases}
`↔` \begin{cases} \left[ \begin{array}{l}\left \{ {{x-2018\ge0} \atop {2020-x\ge0}} \right.\\\left \{ {{x-2018\le0} \atop {2020-x\le0}} \right.\end{array} \right. \\ x=2019 \\ y=2021 \end{cases}
`↔` \begin{cases} \left[ \begin{array}{l}\left \{ {{x\ge2018} \atop {x\le2020}} \right.\\\left \{ {{x\le2018} \atop {x\ge2020}} \right.\end{array} \right. \\ x=2019 \\ y=2021 \end{cases}
`↔` \begin{cases} 2018\le x \le2020 \\ x=2019 \\ y=2021 \end{cases}
`↔ (x;y)=(2019;2021)`