a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

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a) 

PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)

\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)

b) 

PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)

\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)

Vậy pt vô nghiệm.

Ta có : 

\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)

\(\Leftrightarrow\)\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)

\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)

\(\Leftrightarrow\)\(\left(x-2014\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)

Nên \(x-2014=0\)

\(\Rightarrow\)\(x=2014\)

Vậy \(x=2014\)

Chúc bạn học tốt ~ 

\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)

Trừ cả 2 vế cho 2 ta được :

\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)

\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)

\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)

\(\Leftrightarrow\left(x-2014\right)\times\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Mà : \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+17}{1993}\\ \Leftrightarrow\frac{x+1}{2009}+1+\frac{x+3}{2007}+1=\frac{x+5}{2005}+1+\frac{x+17}{1993}\\ \Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}-\frac{x+2010}{2005}-\frac{x+2010}{1993}=0\\ \Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{1993}\right)=0\\ \Leftrightarrow x+2010=0\\ \Leftrightarrow x=-2010\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-2010\right\}\)

\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{1993}\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+3}{2007}+1=\frac{x+5}{2005}+1+\frac{x+7}{1993}+1\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{1993}\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}-\frac{x+2010}{2005}-\frac{x+2010}{1993}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x-2010=0\Leftrightarrow x=2010\)

Vậy \(x=2010\)

\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)'

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)

\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\left(vì:\frac{1}{2009}+\frac{1}{2007}< \frac{1}{2005}+\frac{1}{2003}\right)\)

\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)

\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)

\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)

                                    |________________A________________|

Do A > 0

nên x - 2014 = 0

<=> x = 2014

=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009

=2009

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)

<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)

<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0

<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0

<=> x+2011=0

<=> x=-2011

Vậy pt có nghiệm là x=-2011

\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)

\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}-4=0\)

\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-7}{2007}-1\right)+\left(\dfrac{x-9}{2005}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2007}+\dfrac{x-2014}{2005}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\right)=0\)

\(\Leftrightarrow x-2014=0\) ( do \(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\ne0\))

\(\Leftrightarrow x=2014\)

Vậy phương trình có nghiệm S=\(\left\{2014\right\}\)