\(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)
\(A=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)
\(A=1-\frac{1}{\left|x-2017\right|+2019}\)
A nhỏ nhất khi \(1-\frac{1}{\left|x-2017\right|+2019}\)nhỏ nhất
khi \(\frac{1}{\left|x-2017\right|+2019}\)lớn nhất
khi \(\left|x-2017\right|+2019\)nhỏ nhất
mà |x - 2017| \(\ge0\)
=> |x - 2017| + 2019 \(\ge2019\)
Vậy A nhỏ nhất khi A = 2019 khi x - 2017 = 0 => x = 2017
\(A=\frac{\backslash x-2017\backslash+2018}{\backslash x-2017\backslash+2019}\)
\(A=\frac{2018}{2019}\)
Ta có : \(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)
\(=1-\frac{1}{\left|x-2017\right|+2019}\)
Ta có : \(\left|x-2017\right|\ge0\)
\(\Rightarrow\left|x-2017\right|+2019\ge2019\)
\(\Rightarrow\frac{1}{\left|x-2017\right|+2019}\le\frac{1}{2019}\)
\(\Rightarrow-\frac{1}{\left|x-2017\right|+2019}\ge-\frac{1}{2019}\)
\(\Rightarrow1-\frac{1}{\left|x-2017\right|+2019}\ge1-\frac{1}{2019}=\frac{2018}{2019}\)
Hay : \(A\ge\frac{2018}{2019}\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x-2017\right|=0\Leftrightarrow x=2017\)
Vậy : min \(A=\frac{2018}{2019}\) tại \(x=2017\)
tui có nhầm vài chỗ mấy dòng cuối, mong bạn tha lỗi :((
- Ta có: \(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)
\(\Leftrightarrow A=\frac{\left(\left|x-2017\right|+2019\right)-1}{\left|x-2017\right|+2019}\)
\(\Leftrightarrow A=1-\frac{1}{\left|x-2017\right|+2019}\)
- Để \(A_{min}\)thì \(\frac{1}{\left|x-2017\right|+2019}\)lớn nhất mà \(1>0\)
\(\Rightarrow\left|x-2017\right|+2019\)nhỏ nhất
- Đặt \(B=\left|x-2017\right|+2019\)
- Ta lại có: \(\left|x-2017\right|\ge0 \forall x\)
\(\Rightarrow\left|x-2017\right|+2019\ge2019 \forall x\)
\(\Rightarrow B_{min}=2019\)
\(\Rightarrow A_{min}=1-\frac{1}{2019}=\frac{2018}{2019}\)
- Dấu "=" xảy ra khi \(\left|x-2017\right|=0\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
Vậy \(A_{min}=\frac{2018}{2019} \Leftrightarrow x=2017\)
