+ Nếu x < 2006 thì: A = - x + 2006 + 2007 – x = - 2x + 4013
Khi đó: - x > -2006 => - 2x + 4013 > – 4012 + 4013 = 1 => A > 1
+ Nếu 2006
x
2007 thì: A = x – 2006 + 2007 – x = 1
+ Nếu x > 2007 thì A = x - 2006 - 2007 + x = 2x – 4013
Do x > 2007 => 2x – 4013 > 4014 – 4013 = 1 => A > 1.
Vậy A đạt giá trị nhỏ nhất là 1 khi 2006
x
2007
Bạn kia làm sai đề rồi.
\(A=\left|x-2016\right|+\left|2017-x\right|\)
\(A=\left|x-2016\right|+\left|2017-x\right|\ge\left|x-2016+2017-x\right|\)
Dấu '' = '' xảy ra khi:
\(\left(x-2016\right).\left(2017-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2016\ge0\\2017-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2016\le0\\2017-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2016\\x\le2017\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2016\\x\ge2017\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2016\le x\le2017\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=1\) khi \(2016\le x\le2017.\)
Chúc bạn học tốt!
A = | x - 2016| + | 2017 - x|
\(\Rightarrow A=\left|x-2016\right|+\left|2017-x\right|\ge\left|x-2016+2017-x\right|\)
\(\Rightarrow A\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2016\right)\left(2017-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2016=0\\2017-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2016\\x=2017\end{matrix}\right.\)
Vậy Min A = 1 \(\Leftrightarrow x=2016\) hoặc \(x=2017\)
Học tốt
# Chiyuki Fujito
