Question

Giải phương trình:

a)\(\dfrac{x-2}{3}\) - \(\dfrac{x-3}{4}\)= 1

b)\(\dfrac{x-1}{2015}\) + \(\dfrac{x-2}{2014}+\dfrac{x-5}{2011}+\dfrac{x+1}{2017}\)= 4

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

Answer 1

a) 4( x - 2 ) - 3 ( x - 3 ) = 1

4x - 8 - 3x + 9 =1

x = 0

Answer 2

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 \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\ne0\) )

⇔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=\(\left\{\dfrac{-9}{4}\right\}\)