a)$\dfrac{x-2}{3}-\dfrac{x-3}{4}=1\Leftrightarrow\dfrac{4x-8-3x+9}{12}=1$ ⇔x+1=12⇔x=11 Vậy phương trình đã cho có tập nghiệm S=$\left\{11\right\}$ b)$\dfrac{x-1}{2015}+\dfrac{x-2}{2014}+\dfrac{x-5}{2011}+\dfrac{x+1}{2017}=4$ $\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+\left(\dfrac{x-5}{2011}-1\right)+\left(\dfrac{x+1}{2017}-1\right)=4-4$ $\Leftrightarrow\dfrac{x-1-2015}{2015}+\dfrac{x-2-2014}{2014}+\dfrac{x-5-2011}{2011}+\dfrac{x+1-2017}{2017}=0$ $\Leftrightarrow\dfrac{x-2016}{2015}+\dfrac{x-2016}{2014}+\dfrac{x-2016}{2011}+\dfrac{x-2016}{2017}=0$

$\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\right)=0$

$\Leftrightarrow x-2016=0$ (vì $\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}e0$ )

⇔x=2016

Vậy phương trình đã cho có tập nghiệm S=$\left\{2016\right\}$

c)3(x-1)-5(x+4)+6(2-x)=7 ⇔3x-3-5x-20+12-6x=7⇔3x-5x-6x=7-12+20+3⇔-8x=18⇔$x=\dfrac{-9}{4}$

Vậy phương trình đã cho có tập nghiệm S=$\left\{\dfrac{-9}{4}\right\}$

a, x-2/3-x-3/4=1

⇒(x-2)4=(x+1)3

⇒4x-8=3x+3

⇒x=11

b,ta có :

(x-1/2015-1)+(x-2/2014-1)+(x-5/2011-1)+(x+1/2017-1)=4-4

⇒x-2016/2015+x-2016/2014+x-2016/2011+x-2016/2017=0

⇒(x-2016)(2015+2014+2011+2017)=0

⇒x-2016=0

⇒x=2016

c,3(x-1)-5(x+4)+6(2-x)=7

⇒3x-3-5x-20+12-6x=7

⇒-8x-11=7

⇒x=-9/4