Tìm GTNN của : $A= $ $║x – 2019 ║+ ║ x – 2018 ║+ 2017$ P/s : ║ là dấu gttđ 02/09/2021 Bởi Madelyn Tìm GTNN của : $A= $ $║x – 2019 ║+ ║ x – 2018 ║+ 2017$ P/s : ║ là dấu gttđ
Đáp án: Áp dụng : $|a|$ + $|b|$ $≥$ $|a+b|$ dấu `” =”` khi $a.b ≥0$ $⇒$ $A = |x-2019|+|x-2018| + 2017 ≥ |x-2019|+|2018-x| + 2017 = |x-2019+2018-x| + 2017 = 2018$ Để `” = “` xảy ra thì : `(x-2019).(2018-x)` $≥$ $0$ $TH1$ . $\left \{ {{x-2019 > 0} \atop {2018-x > 0}} \right.$ $⇔$ $\left \{ {{x> 2019} \atop {x<2018}} \right.$ ($KTM$) $TH2$ . $\left \{ {{x-2019 < 0} \atop {2018-x < 0}} \right.$ $⇔$ $\left \{ {{x< 2019} \atop {x>2018}} \right.$ $⇔$ $2018 < x < 2019$ ($TM$) $TH3$ . $\left \{ {{x-2019 = 0} \atop {2018-x = 0}} \right.$ $⇔$ $\left \{ {{x= 2019} \atop {x=2018}} \right.$ ($TM$) Vậy $A$ đạt $GTNN=2018$ khi $2018 ≤ x ≤ 2019$ Bình luận
Đáp án: \[A_{\min}=2018 ⇔ 2018\le x\le 2019\] Giải thích các bước giải: Ta có: \(\begin{split}&\left|x-2019\right|+\left|x-2018\right|\\ &=\left|x-2019\right|+\left|2018-x\right|\ge \left|x-2019+2018-x\right|=1\end{split}\) \(\to A=|x-2019|+|x-2018|+2017\ge 2018\) \(\to \min A=2018\) Dấu “=” xảy ra\(⇔(x-2019)(2018-x)\ge 0\to 2018\le x\le 2019\) Vậy \(A_{\min}=2018 ⇔ 2018\le x\le 2019\) Bình luận
Đáp án:
Áp dụng : $|a|$ + $|b|$ $≥$ $|a+b|$ dấu `” =”` khi $a.b ≥0$
$⇒$ $A = |x-2019|+|x-2018| + 2017 ≥ |x-2019|+|2018-x| + 2017 = |x-2019+2018-x| + 2017 = 2018$
Để `” = “` xảy ra thì : `(x-2019).(2018-x)` $≥$ $0$
$TH1$ . $\left \{ {{x-2019 > 0} \atop {2018-x > 0}} \right.$ $⇔$ $\left \{ {{x> 2019} \atop {x<2018}} \right.$ ($KTM$)
$TH2$ . $\left \{ {{x-2019 < 0} \atop {2018-x < 0}} \right.$ $⇔$ $\left \{ {{x< 2019} \atop {x>2018}} \right.$ $⇔$ $2018 < x < 2019$ ($TM$)
$TH3$ . $\left \{ {{x-2019 = 0} \atop {2018-x = 0}} \right.$ $⇔$ $\left \{ {{x= 2019} \atop {x=2018}} \right.$ ($TM$)
Vậy $A$ đạt $GTNN=2018$ khi $2018 ≤ x ≤ 2019$
Đáp án:
\[A_{\min}=2018 ⇔ 2018\le x\le 2019\]
Giải thích các bước giải:
Ta có: \(\begin{split}&\left|x-2019\right|+\left|x-2018\right|\\ &=\left|x-2019\right|+\left|2018-x\right|\ge \left|x-2019+2018-x\right|=1\end{split}\)
\(\to A=|x-2019|+|x-2018|+2017\ge 2018\)
\(\to \min A=2018\)
Dấu “=” xảy ra\(⇔(x-2019)(2018-x)\ge 0\to 2018\le x\le 2019\)
Vậy \(A_{\min}=2018 ⇔ 2018\le x\le 2019\)