Question

Tìm giá trị nhỏ nhất của:

D=\(\left|x+\dfrac{1}{2}\right|-2017\ge-2017\)

Answer 1

\(\left|x+\dfrac{1}{2}\right|-2017\ge-2017\)

Mà \(\left|x+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow Min_D=-2017\)

Khi \(Min_D=-2017\) thì \(\left|x+\dfrac{1}{2}\right|=0\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy \(Min_D=-2017\) khi \(x=-\dfrac{1}{2}\)

Answer 2

\(D=\left|x+\dfrac{1}{2}\right|-2017\)

\(\left|x+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow D=\left|x+\dfrac{1}{2}\right|-2017\ge-2017\)

Dấu "=" xảy ra khi:

\(\left|x+\dfrac{1}{2}\right|=0\Rightarrow x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)

Vậy \(MIN_D=-2017\)khi \(x=-\dfrac{1}{2}\)