Ta co lx-2016l=l2016-xl>=2016-x

                         lx-1l>=x-1

=>A>= 2015 

Dau bang xay ra khi x-2016<=0

                                   x-1>=0

=>1<=x<=2016

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wopdjoqwedi

Ta có:

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

Vì \(\left|x-2016\right|\ge0\Rightarrow\left|x-2016\right|+2018\ge2018\Rightarrow\frac{1}{\left|x-2016\right|+2018}\le\frac{1}{2018}\)

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

=>\(A_{min}=\frac{2017}{2018}\)<=>|x-2016|=0<=>x-2016=0<=>x=2016

Áp dụng : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Ta có : \(A=\left|x-2016\right|+\left|x-1\right|=\left|x-2016\right|+\left|1-x\right|\ge\left|x-2016+1-x\right|\)

                \(=\left|2017\right|=2017\)

\(\Rightarrow Min_A=2017\)

Áp dụng:\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Ta có:A=\(\left|x-2016\right|+\left|x-1\right|=\left|x-2016\right|+\left|1-x\right|\ge\left|x-2016+1-x\right|=\left|2017\right|=2017\)

\(\Rightarrow GTNN\) của A là:2017

Có \(\left|x-2016\right|=\left|2016-x\right|\)

\(\Rightarrow A=\left|2016-x\right|+\left|x-1\right|\ge\left|2016-x+x-1\right|\)

\(\Rightarrow A\ge2015\)

\(\Rightarrow Min_A=2015\Leftrightarrow\left(2016-x\right)\left(x-1\right)\ge0\)

Ta có bảng xét dấu :

\(\Rightarrow1\le x\le2016\)

Vậy ...

Ta có : A = l2014 - x l + l 2015 - x l + l2016 - x l 
        => A = l2014 - x l + l2015 - x l + l x-2016 l   (Với x>2016 )
         => A >= l 2014 -x + x- 2016 l + l2015 -x l
        => A >= l2014-2016l + l2015-x l
       => A >= l -2 l + l2015 - x l
        => A >= 2 + l2015 - x l 
      Vì l2015 - x l >=0 Nên <=> A >= 2 +0
                                         => A >=2 
  Vậy Min A =2 <=> l2015 - x l = 0 
                         => 2015 - x= 0   => x= 2015-0 =2015
Vậy tại x= 2015 thì GTNN của A =2 

sai rồi

Vì \(\left|x-7\right|\ge0;\left|x-2016\right|\ge0;\left|x-2017\right|\ge0\)

         Suy ra:\(\left|x-7\right|+\left|x+2016\right|+\left|x-2017\right|\ge0\)

      Dấu = xảy ra khi x-7=0;x=7

                                 x+2016=0;x=-2016

                                 x-2017=0;x=2017

Vậy Min A=0 khi x=7;-2016;2017

A = |x-7|+|x-2016|+|x-2017|

= |x-7|+|x-2016|+|2017-x|

≥ |x-7+2017-x|+|x-2016| = 2017+|x-2016|≥2017

để A nhỏ nhất => A = 2017

=> |x - 2016| = 0 => x = 2016