\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)

Vậy x=-2015

\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)

\(\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-1}{1008}-2\right)+\left(\dfrac{x}{2017}-1\right)\)

\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0\)

\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)

\(x-2017=0\) vì\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)

\(\Rightarrow x=2017\)

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)

\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)

\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)

\(\Leftrightarrow x=2017\)         \(\text{Vậy }x=2017\)

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)

\(\left(\dfrac{x+1}{2015}+1\right)+\left(\dfrac{x+2}{2014}+1\right)+\left(\dfrac{x+3}{2013}+1\right)=0\)

\(\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)

\(\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)

\(\Rightarrow x+2016=0\Rightarrow x=-2016\)

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)

\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1+\dfrac{x+3}{2013}+1=0\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)

\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)

\(\Rightarrow x+2016=0\Rightarrow x=-2016\)

Chúc bạn học tốt!!!

\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\dfrac{x+4}{2012}+1+\dfrac{x+3}{2013}+1=\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\left(\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\right)=0\)

\(\Leftrightarrow x+2016.\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)

Vì \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

Vậy \(x=-2016\) tại biểu thức \(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

Theo đề ta có: x+4/2012+x+3/2013=x+2/2014+x+1/2015
=>x+4/2012+x+3/2013-x+2/2014+x+1/2015=0
=>( x+4/2012+1)+(x+3/2013+1)-(x+2/2014+1)+(x+1/2015+1)
=>x+2016/2012+x+2016/2013-x+2016/2014-x+2016/2015=0
=>x+2016.(1/2012+1/2013-1/2014-1/2015)=0
Do 1/2012+1/2013-1/2014-1/2015>0
nên x+2016=0
=>x=-2016
Vậy x=-2016

\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2012}+1\right)+\left(\dfrac{x+3}{2013}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2014}+1\right)\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\dfrac{x+2016}{2014}-\dfrac{x+2016}{2015}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)

Vì \(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\ne0\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}+\dfrac{x+2024}{2}=0\)

\(\Leftrightarrow(\dfrac{x+1}{2015}+1)+(\dfrac{x+2}{2014}+1)+(\dfrac{x+3}{2013}+1)+(\dfrac{x+4}{2012}+1)+\dfrac{x+2024}{2}-4=0\)\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}+\dfrac{x+2016}{2}=0\)\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}\right)=0\)

Hiển nhiên: \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}>0\)

\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)

x =0 nhé bn

\(\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)

<=>\(\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)

<=>(x+2017)(1/2016+1/2015-1/2014-1/2013)=0

vì 1/2016+1/2015-1/2014-1/2013 khác 0

nên x+2017=0<=>x=-2017

vậy................

chúc bạn học tốt ^^

. Đây nha