Question

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

Answer 1

Đặt: \(\left|x-2017\right|=t\ge0\) ta có: \(l=\frac{t+2017}{t+2018}=\frac{t+2018-1}{t+2018}=1-\frac{1}{t+2018}\ge1-\frac{1}{2018}=\frac{2017}{2018}\)

Dấu "=" xảy ra khi: \(t=0\Leftrightarrow x=2017\)

Answer 2

Đặt: |x−2017|=t≥0 ta có: l=t+2017t+2018 =t+2018−1t+2018 =1−1t+2018 ≥1−12018 =20172018 

Dấu "=" xảy ra khi: t=0⇔x=2017

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Answer 3

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

A bé nhất khi \(\frac{1}{\left|x-2017\right|+2018}\) lớn nhất.

Mà \(\frac{1}{\left|x-2018\right|+2018}\le\frac{1}{2018}\forall x\) (do \(\left|x-2018\right|\ge0\forall x\))

Suy ra \(A\ge1-\frac{1}{2018}=\frac{2017}{2018}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|x-2017\right|=0\Leftrightarrow x=2017\)

Vậy \(A_{min}=\frac{2017}{2018}\Leftrightarrow x=2017\)