(3x+1)2-4(x-3)2=0
=9x2+6x+1-4(x2-6x+9)=0
=9x2+6x+1-4x2+24x-36=0
=5x2+30x-35=0
Vậy x=1 và-7
\(\left(3x+1\right)^2-4\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(3x\right)^2+2.3x+1-4\left(x^2-2.x.3+3^2\right)=0\)
\(\Leftrightarrow9x^2+6x+1-4x^2+24x-36=0\)
\(\Leftrightarrow5x^2+30x-35=0\)
\(\Leftrightarrow5\left(x^2+2.x.3+3^2\right)-80=0\)
\(\Leftrightarrow5\left(x+3\right)^2=80\)
\(\Leftrightarrow\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy ...
(3x + 1)2 - 4(x - 3)2 = 0
(3x + 1)2 - [2(x - 3)]2 = 0
(3x + 1 + 2x - 6)(3x + 1 -2x + 6) = 0
(5x - 5)(x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}5x-5=0\\x+7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x=5\\x=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy x = 1 ;x = -7
