Tìm giá trị nhỏ nhất của:P=/x-2016/+/x-2017/.

Áp dụng BĐT /a+b/. ≤/a/+/b/. ⇒ P=/x-2016/+/x-2017/= /x-2016/+/2017-x/ lớn hơn hoặc bằng /x-2016+2017-x/=1.

Vậy GTNN của P là 1 <=> 0. ≤(x-2016)(2017-x) <=> 2016. ≤x. ≤2017. 

\(A=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)

\(A=\frac{\left|x-2017\right|+2019-1}{\left|x-2017\right|+2019}\)

\(A=1-\frac{1}{\left|x-2017\right|+2019}\)

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 \(x=2017\)

\(A=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|+\left|x-2019\right|=\left|x-2016\right|+\left|x-2017\right|+\left|2018-x\right|+\left|2019-x\right|\ge\left|x-2016+x-2017+2018-x+2019-x\right|=4\)

A=|x−2016|+|x−2017|+|x−2018|+|x−2019|=|x−2016|+|x−2017|+|2018−x|+|2019−x|≥|x−2016+x−2017+2018−x+2019−x|=4A=|x−2016|+|x−2017|+|x−2018|+|x−2019|=|x−2016|+|x−2017|+|2018−x|+|2019−x|≥|x−2016+x−2017+2018−x+2019−x|=4

\(C=\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\)

\(=1-\frac{1}{\left|x-2017\right|+2019}\)

Vì \(\left|x-2017\right|\ge0;\forall x\)

\(\Rightarrow\left|x-2017\right|+2019\ge2019;\forall x\)

\(\Rightarrow\frac{1}{\left|x-2017\right|+2019}\le\frac{1}{2019};\forall x\)

\(\Rightarrow-\frac{1}{\left|x-2017\right|+2019}\ge-\frac{1}{2019};\forall x\)

\(\Rightarrow1-\frac{1}{\left|x-2017\right|+2019}\ge\frac{2018}{2019};\forall x\)

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

                     \(\Leftrightarrow x=2017\)

Vậy \(C_{min}=\frac{2018}{2019}\)\(\Leftrightarrow x=2017\)

THANKS BẠN NHA

M = | x - 2019 | + | x - 2018 | - 2017

M = | x - 2019 | + | x - 2018 | - 2017 \(\ge\)- 2017

Dấu " = " xảy ra \(\Leftrightarrow\)x - 2019 = 0 hoặc x - 2018 = 0

\(\Rightarrow\)x = 2019 hoặc x = 2018

Min M = - 2017 \(\Leftrightarrow\)x = 2019 hoặc x = 2018

*) Ta chứng minh bổ đề: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2\ge\left(\left|a+b\right|\right)^2\)

\(\Leftrightarrow a^2+b^2+2\left|ab\right|\ge a^2+b^2+2ab\)

\(\Leftrightarrow2\left|ab\right|\ge2ab\) 

\(\Leftrightarrow\left|ab\right|\ge ab\) ( luôn đúng ) 

Dấu "=" xảy ra khi \(ab\ge0\)

Theo bài cho: M = |x-2019| + |x-2018| - 2017

=> M = |x - 2019| + |2018 - x| - 2017

Áp dụng bổ đề trên => | x - 2019 | + | 2018 - x| \(\ge\) | x - 2019 + 2018 - x |

=> | x - 2019 | + | 2018 - x | \(\ge\)1

=> | x - 2019 | + | 2018 - x | - 2017 \(\ge\)1 - 2017

=> M \(\ge\)-2016

Dấu "=" xảy ra khi ( x - 2019 ).( 2018 - x)\(\ge\)0

Ta xét 2 trường hợp:

+) Nếu \(\hept{\begin{cases}x-2019\ge0\\2018-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2019\\x\le2018\end{cases}}\)( loại )

+) Nếu \(\hept{\begin{cases}x-2019\le0\\2018-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2019\\x\ge2018\end{cases}}\)\(\Leftrightarrow2018\le x\le2019\)( thỏa mãn )

Vạy: GTNN của M = -2016 khi \(2018\le x\le2019\)