a, Ta có :
\(A=\left|2x-2\right|+\left|2x-2017\right|=\left|2x-2\right|+\left|2017-2x\right|\ge\left|2x-2+2017-2x\right|=2015\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x-2\right)\left(2017-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-2\ge0\\2017-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-2\le0\\2017-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x\ge2\\2017\ge2x\end{matrix}\right.\\\left\{{}\begin{matrix}2x\le2\\2017\le2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\\dfrac{2017}{2}\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\\dfrac{2017}{2}\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1\le x\le\dfrac{2017}{2}\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
b, Tương tự
c, \(\left|x+3\right|+\left|x+7\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+7\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x-3\right|+\left|x+7\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(x\ge0\) ta có :
+) \(\left|x+3\right|=x+3\)
\(\left|x+7\right|=x+7\)
\(\Leftrightarrow\left|x+3\right|+\left|x+7\right|=x+3+x+7=4x\)
\(\Leftrightarrow2x+10=4x\)
\(\Leftrightarrow10=2x\)
\(\Leftrightarrow x=5\)
Vậy ..
B1b)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2\right|+\left|x-8\right|\)
\(B\ge\left|x-2\right|+\left|8-x\right|=6\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(8-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\le0\\8-x\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\ge0\\8-x\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2\\x\ge8\end{matrix}\right.\left(C\right)}\\\left\{{}\begin{matrix}x\ge2\\x\le8\end{matrix}\right.\left(L\right)}\end{matrix}\right.\)
TH1: chọn, TH2: loại.
Vậy \(MIN_B=6\Leftrightarrow2\le x\le8\)
