Tìm `A_{min}` biết : `A = |x – 2019| + |x – 2020| + |y – 2021| + |x – 2022| + 2016` 07/10/2021 Bởi Brielle Tìm `A_{min}` biết : `A = |x – 2019| + |x – 2020| + |y – 2021| + |x – 2022| + 2016`
`A=//x-2019//+//x-2020//+//y-2021//+//x-2022//+2016` \(\left[ \begin{array}{l}x-2019≥0\\x-2020≥0\\y-2021≥0\\x-2022≥0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x≥2019\\x≥2020\\y≥2021\\x≥2022\end{array} \right.\) ĐỂ A có min ⇔\(\left[ \begin{array}{l}x=2019\\x=2020\\y=2021\\x=2022\end{array} \right.\) `⇒với x=2019 ` `⇒y=2021` `⇒minA=1+2+2016=2019` `⇒với x=2020 ` `⇒y=2021` `⇒minA=1+2+2016=2019` `⇒với x=2022 ` `⇒y=2021` `⇒minA=1+2+2016=2019` `⇒minA=2019 khi (x;y)∈(2019;2021);(2020;2021);(2021;2021)` Bình luận
Đáp án: Giải thích các bước giải: Ta có : `A = |x – 2019| + |x – 2020| + |y – 2021| + |x – 2022| + 2016` `=> A = |x-2019|+|2022-x|+|x-2020|+|y-2021|+2016` `=> A ≥ |x-2019+2022-x|+0+0+2016` `=> A ≥ 2019` Vậy `A_min=2019` $⇔ \left\{\begin{matrix} (x-2019)(2022-x)≥0& \\ y-2021=0& \\ x-2020=0& \end{matrix}\right.$ $⇔ \left\{\begin{matrix} (x-2019)(2022-x)≥0& \\ y=2021& \\ x=2020& \end{matrix}\right.$ Bình luận
`A=//x-2019//+//x-2020//+//y-2021//+//x-2022//+2016`
\(\left[ \begin{array}{l}x-2019≥0\\x-2020≥0\\y-2021≥0\\x-2022≥0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x≥2019\\x≥2020\\y≥2021\\x≥2022\end{array} \right.\)
ĐỂ A có min
⇔\(\left[ \begin{array}{l}x=2019\\x=2020\\y=2021\\x=2022\end{array} \right.\)
`⇒với x=2019 `
`⇒y=2021`
`⇒minA=1+2+2016=2019`
`⇒với x=2020 `
`⇒y=2021`
`⇒minA=1+2+2016=2019`
`⇒với x=2022 `
`⇒y=2021`
`⇒minA=1+2+2016=2019`
`⇒minA=2019 khi (x;y)∈(2019;2021);(2020;2021);(2021;2021)`
Đáp án:
Giải thích các bước giải:
Ta có :
`A = |x – 2019| + |x – 2020| + |y – 2021| + |x – 2022| + 2016`
`=> A = |x-2019|+|2022-x|+|x-2020|+|y-2021|+2016`
`=> A ≥ |x-2019+2022-x|+0+0+2016`
`=> A ≥ 2019`
Vậy `A_min=2019` $⇔ \left\{\begin{matrix} (x-2019)(2022-x)≥0& \\ y-2021=0& \\ x-2020=0& \end{matrix}\right.$
$⇔ \left\{\begin{matrix} (x-2019)(2022-x)≥0& \\ y=2021& \\ x=2020& \end{matrix}\right.$