\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(4x^2-4x+1-\left(x^2+6x+9\right)=0\)
\(4x^2-4x+1-x^2-6x-9=0\)
\(4x^2-x^2-4x-6x+1-9=0\)
\(3x^2-10x-8=0\)
\(3x^2-12x+2x-8=0\)
\(3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\left(x-4\right).\left(3x+2\right)=0\)
=> \(x-4=0\) hay \(3x+2=0\)
=> \(x=4\) hay \(3x=-2\)
=> \(x=4\) hay \(x=-\dfrac{3}{2}\)
Vậy \(x=4\) hay \(x=-\dfrac{3}{2}\)
