<=>(2x)2-(x+3)2=0
<=>(2x-x-3)(2x+x+3)=0
<=>-3x.3(x+1)=-9x.(x+1)=0
=>x=0 hoặc x+1=0
<=>x=-1
4x2 - (x+3)2 = 0
<=> 4x2 - (x2+6x+9) = 0
<=> 4x2 - x2 - 6x - 9 = 0
<=>3x2 - 6x = 9
<=> x(3x-6) = 9
Lập bảng a.b = 9 ra được: x \(\in\){3;-1}
\(4x^2-\left(x+3\right)^2=0\\ \left(2x-x-3\right)\left(2x+x+3\right)=0\\ \left(x-3\right)\left(3x+3\right)=0\\ \left\{{}\begin{matrix}x-3=0\\3x+3=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)