=>2x^2-7x+3-2x^2-2x=0
=>-9x=-3
=>x=1/3
\(\Leftrightarrow2x^2-6x-x+3-2x^2-2x=0\\ \Leftrightarrow-9x=-3\\\Leftrightarrow x=\dfrac{-3}{-9}=\dfrac{1}{3} \)
Vậy \(x=\dfrac{1}{3}\)
\(\left(2x-1\right)\left(x-3\right)-2x\left(x+1\right)=0\\ x\left(2x-1\right)-3\left(2x-1\right)-2x^2-2x=0\\ 2x^2-x-6x+3-2x^2-2x=0\\ \left(2x^2-2x^2\right)-\left(x+6x+2x\right)+3=0\\ 3-9x=0\\ 9x=3\\ x=\dfrac{1}{3}\)