\(\left(3x-7\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^2-2.3x.7+7^2-4\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow9x^2-42x+49-4x^2-8x-4=0\)
\(\Leftrightarrow5x^2-50x+45=0\)
\(\Leftrightarrow x^2-10x+9=0\)
\(\Leftrightarrow x^2-x-9x+9=0\)
\(\Leftrightarrow x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy................
Ta có: \(\left(3x-7\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\)
\(\Leftrightarrow5\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{1;9\right\}\)