\(\dfrac{x+2014}{2}+\dfrac{2\left(x+2014\right)}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)

\(\left(x+2014\right)\left(\dfrac{1}{2}+\dfrac{2}{7}\right)=\left(x+2014\right)\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)

\(\left(x+2014\right)\dfrac{11}{14}=\left(x+2014\right)\dfrac{11}{30}\)

Dấu ''=''↔x=-2014

\(\dfrac{x-2014}{4}+\dfrac{x-2015}{3}=\dfrac{x-13}{2005}+\dfrac{x-14}{2004}\)

<=>\(\left(\dfrac{x-2014}{4}-1\right)+\left(\dfrac{x-2015}{3}-1\right)=\left(\dfrac{x-13}{2005}-1\right)+\left(\dfrac{x-14}{2004}-1\right)\)

<=>\(\dfrac{x-2018}{4}+\dfrac{x-2018}{3}=\dfrac{x-2018}{2005}+\dfrac{x-2018}{2004}\)

<=>\(\left(x-2018\right).\left[\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2005}-\dfrac{1}{2004}\right]=0\)

<=>  \(x-2018=0\)

=>x=2018

Vậy S= {2018}

Chúc bạn học tốt!
#Yuii

thank

a: =>4x-6-9=5-3x-3

=>4x-15=-3x+2

=>7x=17

hay x=17/7

b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)

=>2/3x+21/3x=4/5+2+1/4=61/20

=>23/3x=61/20

=>3x=23:61/20=460/61

hay x=460/183

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

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

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

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

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

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

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

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

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

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

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

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

Ta có : \(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}=4\)

\(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}-4=0\)

\(\dfrac{x-3}{2015}-1+\dfrac{x-4}{2014}-1+\dfrac{x-5}{2013}-1+\dfrac{x-6}{2012}-1=0\)

\(\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}+\dfrac{x-2018}{2013}+\dfrac{x-2018}{2012}=0\)

\(\left(x-2018\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}\right)=0\)

Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}>0\)

=> x - 2018 = 0

x = 0 + 2018

x = 2018

Vậy x = 2018

a) \(5x-3=7\)

\(\Leftrightarrow5x=7+3\)

\(\Leftrightarrow5x=10\)

\(\Leftrightarrow x=\dfrac{10}{5}\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

b) \(\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow x+3=0\) hoặc \(x-4=0\)

*) \(x+3=0\)

\(x=0-3\)

\(x=-3\)

*) \(x-4=0\)

\(x=0+4\)

\(x=4\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\left|x^2+2014\right|=1\)

\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)

*) \(x^2+2014=1\)

\(\Leftrightarrow x^2=1-2014\)

\(\Leftrightarrow x^2=-2013\) (vô lý)

*) \(x^2+2014=-1\)

\(\Leftrightarrow x^2=-1-2014\)

\(\Leftrightarrow x^2=-2015\) (vô lý)

Vậy \(S=\varnothing\)

d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)

ĐKXĐ: \(x\ne-1;x\ne3\)

\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-6-x-1=3x-11\)

\(\Leftrightarrow-2x=-11+7\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\) (nhận)

Vậy \(S=\left\{2\right\}\)