\(\Leftrightarrow\left|x+\frac{5}{2}\right|+\left|\frac{2}{5}-x4\right|=\frac{2\left|5x-2\right|+5\left|2x+5\right|}{10}\)

\(\Rightarrow\frac{2\left|5x-2\right|+5\left|2x+5\right|}{10}=0\)

=>x\(\in\){rỗng} x ko tồn tại với nghiệm số thực

toán của bn dễ ẹc

x = {3, 4 , 4,6,7,8,9,10,.....}

li-ke mình nhé

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

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

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

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

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

x = -2014

X = 2014 

ĐUNG

| x + 2015 | + 2016 = 2017 

| x + 2015 | = 2017 - 2016 = 1 

x + 2015 = 1 hoặc x + 2015 = -1 

x = 1 - 2015 hoặc x = -1 - 2015 

x = -2014 hoặc x = - 2016 

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

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

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

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

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

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

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