\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)

\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)

\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)

\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)

\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)

Vậy ....

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)

Khi đó phương trình trở thành: 

\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)

Tick plz

\(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)

\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}=\dfrac{x+2019}{2017}+\dfrac{x+2019}{2018}\)

\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}-\dfrac{x+2019}{2017}-\dfrac{x+2019}{2018}=0\)

\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)

Mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)

\(\Leftrightarrow x+2019=0\)

\(\Leftrightarrow x=-2019\)

Vậy...

\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)

\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)

\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)

\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)

\(=>\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)

=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)

=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\) = 0

Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)

=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)

=> \(x+2018=0\)

=>x = 0-2018

=> x = 0+(-2018)

=> x = -2018

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)

<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)

<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)

<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)

<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)

vì 1/2015+1/2016-1/2017-1/2018 khác 0

=>x-2014=0<=>x=2014

vậy.....................

chúc bạn học totts ^^

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)

\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)

\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)

\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)

Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)

\(\Rightarrow x-2014=0\)

\(\Rightarrow x=2014\)

Vậy........

Lấy cả hai về mỗi số trừ đi 1

chuyển cả bốn số về 1 vế

chuyen ve (x-2014)(...)

chung minh(...)< or > 0

rồi ra x= 2014

\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)

\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)

\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)

\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)

\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)

Vậy ...