\(\dfrac{3x-2}{4}-\dfrac{2x+1}{3}=\dfrac{x-2}{6}-\dfrac{x+1}{12}\\ \dfrac{3\left(3x-2\right)}{12}-\dfrac{4\left(2x+1\right)}{12}=\dfrac{2\left(x-2\right)}{12}-\dfrac{x+1}{12}\\ 3\left(3x-2\right)-4\left(2x+1\right)=2\left(x-2\right)-\left(x+1\right)\\ 9x-6-8x-4=2x-4-x-1\\ x-10=x-5\\ x-x=-5+10\\ 0=5\)
=> Không có x thỏa mãn
\(\dfrac{3x-2}{4}-\dfrac{2x+1}{3}=\dfrac{x-2}{6}-\dfrac{x+1}{12}\)
=>\(\dfrac{3\left(3x-2\right)-4\left(2x+1\right)}{12}=\dfrac{2\left(x-2\right)-x-1}{12}\)
=>\(3\left(3x-2\right)-4\left(2x+1\right)=2\left(x-2\right)-x-1\)
=>\(9x-6-8x-4=2x-4-x-1\)
=>x-10=x-5
=>-10=-5(vô lý)
Vậy: Phương trình vô nghiệm
