\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+11x+3=0\)
\(\Leftrightarrow7x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{7}\)
Ta có: \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+12x-x+3=0\)
\(\Leftrightarrow7x=-4\)
hay \(x=-\dfrac{4}{7}\)