|x+2015|+2016=2017

|x+2015|=2017-2016

|x+2015|=1

|x+2015|=+1

=>x+2015=1=>x=-2014

=>x+2015=-1=>x=-2016

Vay .....................

k mik nha ban

ta co:/x+2015/=2017-2016=1

{x+2015=1;x=-2014

x+2015=-1;x=-2016

theo đề bài: /x+2015/+2016=2017 (đkxđ: x c  Z)

=> /x+2015/=2017-2016=1

=> x+2015=1 hoặc x+2015=-1

giải hai bài toán ta được tập hợp xc { -2014; -2016}

/x+2014/+2015=2016

=>/x+2014/=2016-2015

=>/x+2014/=1

=>x+2014=1 hoặc x+2014=-1

*nếu x+2014=1

=>            x=1-2014

=>            x=-2013

*nếu x+2014=-1

=>             x=-1-2014

=>             x=-2015

vậy x thuộc {-2015;-2013}

mk thi rùi đúng 100%

|x + 2014| + 2015 = 2016

=> |x + 2014| = 2016 - 2015

=> |x + 2014| = 1

+) x + 2014 = 1

=> x = 1 - 2014

=> x = -2013

+) x + 2014 = -1

=> x = -1 - 2014

=> x = -2015

Vậy S = {-2015; -2013}.

Ta có:|x+2014|+2015=2016

=>|x+2014|=1

TH1:x+2014=-1

=>x=-2015

TH2:x+2014=1

=>x=-2013

Vậy x=-2015;-2013

| x + 2014 | + 2015 = 2016

=> | x + 2014 | = 2016 - 2015

=> | x + 2014 | = 1

=> x + 2014 = 1 hoặc - 1

TH1 : x + 2014 = 1 => x = 1 - 2014 => x = - 2013

TH2 : x + 2014 = - 1 => x = - 1 - 2014 => x = - 2015

Vậy x = -2013 hoặc x = - 2015

\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)

\(\Rightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)

\(\Rightarrow\frac{x}{2015}-\frac{x}{2017}=0\)

\(\Rightarrow x.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)

Mà ta thấy \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow x=0\)

Vậy \(x=0\)

\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)

\(\Leftrightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)

\(\Leftrightarrow x\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)

\(\Leftrightarrow x=0\).Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\ne0\)

Vậy giá trị của x là x=0

\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)

<=>\(\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)=0\)

<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)

<=>\(\left|x\right|.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)

Vì \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow\left|x\right|=0\Rightarrow x=0\)

Vậy x=0

\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)

\(\Rightarrow\left|x.\left(\frac{1}{2015}+\frac{1}{2016}\right)\right|=\left|x.\left(\frac{1}{2016}+\frac{1}{2017}\right)\right|\)

\(\Rightarrow\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)

\(\Rightarrow\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)

Mà \(\frac{1}{2015}+\frac{1}{2016}>\frac{1}{2016}+\frac{1}{2017}\)

=> |x| = 0

=> x = 0

Vậy x = 0

1 so la so -2014

tk nha

ta có: |x+2015|+2016=2017

=>|x+2015| = 1

=> \(\orbr{\begin{cases}x+2015=-1\\x+2015=1\end{cases}}\)=> \(\orbr{\begin{cases}x=-2016\\x=-2014\end{cases}}\)

vậy: \(\orbr{\begin{cases}x=-2016\\x=-2014\end{cases}}\)

bạn yk mk nha, thank.

|x+2015|+2016=2017

|x+2015|=2017-2016

|x+2015|=1

Suy ra x\(\in\){-2014;-2016}

Vậy có 2 số thỏa mãn đề bài.

-2016;2016

x = (-2016; 2016)