1.
Ta có: \(\left|2x-\dfrac{1}{2}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)
\(\Rightarrow A\ge-2017\)
Dấu "=" xảy ra \(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy, MinA = -2017 \(\Leftrightarrow x=\dfrac{1}{4}\)
1) Tìm giá trị nhỏ nhất của biểu thức:
A = \(\left|2x-\dfrac{1}{2}\right|\) - 2017
Ta có:
\(\left|2x-\dfrac{1}{2}\right|\) ≥ 0
=> \(\left|2x-\dfrac{1}{2}\right|\) - 2017 ≥ -2017
Dấu " = " xảy ra khi \(2x-\dfrac{1}{2}\) = 0 hay x = \(\dfrac{1}{4}\)
Vậy Min A = -2017 khi x = \(\dfrac{1}{4}\)
1/ Ta có :
\(\left|2x-\dfrac{1}{2}\right|\ge0\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)
\(\Leftrightarrow A\ge-2017\)
Dấu "=" xảy ra khi :
\(\left|2x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy ....
2/ Ta có :
\(\left|x+2017\right|\ge0\)
\(\Leftrightarrow-\left|x+2017\right|\le0\)
\(\Leftrightarrow2017-\left|x+2017\right|\le2017\)
\(\Leftrightarrow B\le2017\)
Dấu "=" xảy ra khi :
\(\left|x+2017\right|=0\Leftrightarrow x=-2017\)
Vậy..
