Question

[NGUYỄN CẨM TÚ]

Tìm GTNN của biểu thức A= \(\dfrac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}\)

Answer 1

[Nguyễn Huy Tú]

\(A=\dfrac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}=1-\dfrac{1}{\left|x-2016\right|+2018}\)

Để A nhỏ nhất thì \(\dfrac{1}{\left|x-2016\right|+2018}\) lớn nhất thì \(\left|x-2016\right|+2018\) nhỏ nhất

Ta có: \(\left|x-2016\right|\ge0\)

\(\Rightarrow\left|x-2016\right|+2018\ge2018\)

\(\Rightarrow\dfrac{1}{\left|x-2016\right|+2018}\le\dfrac{1}{2018}\)

\(\Rightarrow A=1-\dfrac{1}{\left|x-2016\right|+2018}\ge1-\dfrac{1}{2018}=\dfrac{2017}{2018}\)

Dấu " = " khi \(\left|x-2016\right|=0\Rightarrow x=2016\)

Vậy \(MIN_A=\dfrac{2017}{2018}\) khi x = 2016

Answer 2

[Kirigawa Kazuto]

Ta có :

\(A=\dfrac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}=\dfrac{\left|x-2016\right|+2018-1}{\left|x-2016\right|+2018}=1-\dfrac{1}{\left|x-2016\right|+2018}\)Vì \(\left|x-2016\right|\ge0\Rightarrow\left|x-2016\right|+2018\ge2018\)

\(\Rightarrow\dfrac{1}{\left|x-2016\right|+2018}\le\dfrac{1}{2018}\)

\(\Rightarrow1-\dfrac{1}{\left|x-2016\right|+2018}\ge\dfrac{2017}{2018}\)

\(\Rightarrow A_{min}=\dfrac{2017}{2018}\)

<=> |x - 2016| = 0

<=> x = 2016