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\(\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\ge\left|x-2015\right|+\left|x-2017\right|\ge\left|2015-x+x-2017\right|\ge2\)

đẳng thức xảy ra khi \(2015\le x\le2017\)

Bạn kia sai cmnr nhé:

\(linh=\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\)

\(linh=\left|x-2015\right|+\left|x-2017\right|+\left|x-2016\right|\)

\(linh=\left|x-2015\right|+\left|2017-x\right|+\left|x-2016\right|\)

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

Nên \(linh\ge\left|x-2015+2017-x\right|+\left|x-2016\right|\)

\(linh\ge2+\left|x-2016\right|\) Vì \(\left|x-2016\right|\ge0\) nên

\(linh\ge2\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-2015\ge0\\x-2016=0\\x-2017\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2015\\x=2016\\x\le2017\end{matrix}\right.\)

Nên \(x=2016\)

\(\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\)

\(=\left|x-2015\right|+\left|2017-x\right|+\left|x-2016\right|\)

\(\ge\left|x-2015+2017-x\right|+\left|x-2016\right|\)

\(=2+\left|x-2016\right|\ge2\)

Dấu "=" khi \(\hept{\begin{cases}x-2016=0\\\left(x-2015\right)\left(2017-x\right)\ge0\end{cases}}\Leftrightarrow x=2016\)

a)Vì |x−2015|= 1/2 nên x-2015=-1/2 hoặc x-2015=1/2

Nếu x-2015=-1/2 thì

x=2015+(-1)/2

x=4029/2

Nếu x-2015=1/2 thì

x=2015+1/2

x=4031/2

Vậy x=4029/2

hoặc x=4031/2

b)

Nếu x>2016 thì |x−2015|=x-2015 ,|x−2016|=x-2016

Khi đó: |x−2015|+|x−2016|=2017

=>x-2015+x-2016=2017

=>2x-4031=2017

=>2x=6048=>x=3024(thỏa mãn x>2016)

Nếu 2015<x<2016 thì |x−2015|=x-2015,

|x−2016|=2016-x. khi đó

|x−2015|+|x−2016|=2017

=>x-2015+2016-x=2017

=>1=2017(vô lý loại)

Nếu x>2015 thì |x−2015|=2015-x,|x−2016|=2016-x

Khi đó:

|x−2015|+|x−2016|=2017

=>2015-x+2016-x=2017

=>4031-2x=2017

=>2x=2014=>x=1007(thỏa mãn x<2015)

Vậy x=1007 hoặc x=3024