\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)

\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)

\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)

\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)

ta co (2x-1)(3x+1)+(3x+4)(3-2x)=5

     (=)6x²-3x+2x-1+6x-6x²+12-8x=5

     (=)-4x+11=5

     (=)-4x=-6

     (=)x=3/2

(2x-1)(3x+1)+(3x-4)(3-2x)=5

<=> 6x²+2x-3x-1+9x-6x²-12+8x=5

<=> 16x-13=5

<=> 16x = 18

<=> x=9/8

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

=3/4x+5-2/3x-4-1/6x+1=1/3x+4-1/3x-3

=-1/12=7

x=84

Đ/S...

nhận xét 3 >0

=> /3x+1/= 3x+1

/4-x/= 4-x

ta có 3x+1 + 4 -x = 3

=> 2x + 5 = 3

=> 2x = -2

=> x= -1

cam on ban Nguyen Quang Tung

Giải sai bét 

|3x+1|>4

=>TH1: 3x+1>4

<=>3x>3

<=>x>1

TH2:3x+1<-4

<=>3x<-5

<=>x< -5/3