\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\)

\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))

Cộng 1 vào mỗi số hạng là ra

\(\Rightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\\ \Rightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}=\dfrac{x+2020}{2017}+\dfrac{x+2020}{2016}\\ \Rightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\\ \Rightarrow x=-2020\)

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

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

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

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

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

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

\(\Leftrightarrow x+2019=0\)

\(\Leftrightarrow x=-2019\)

Vậy...

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

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

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

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

2019 -x =0 ; x =2019

Giải:

\(\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)

\(\Leftrightarrow2+\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=2+\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)

\(\Leftrightarrow\dfrac{x+2015}{5}+1+\dfrac{x+2016}{4}+1=\dfrac{x+2017}{3}+1+\dfrac{x+2018}{2}+1\)

\(\Leftrightarrow\dfrac{x+2015+5}{5}+\dfrac{x+2016+4}{4}=\dfrac{x+2017+3}{3}+\dfrac{x+2018+2}{2}\)

\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}=\dfrac{x+2020}{3}+\dfrac{x+2020}{2}\)

\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}-\dfrac{x+2020}{3}-\dfrac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

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

\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}=\dfrac{x-3}{2017}+\dfrac{x-4}{2016}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)=\left(\dfrac{x-3}{2017}-1\right)+\left(\dfrac{x-4}{2016}-1\right)\)

\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}-\dfrac{x-2020}{2017}-\dfrac{x-2010}{2016}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)

\(\Rightarrow x-2020=0\Leftrightarrow x=2020\)

vậy.......

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

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

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

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

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

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

a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)

\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)

\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)

Vậy A < B

b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)

\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)

\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)

Vậy M < N

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

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

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

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