\(\left(3x-1\right)\left(x-3\right)+\left(x^2-9\right)=0\\ \left(x-3\right)\left(3x-1+x+3\right)=0\\ \left(x-3\right)\left(4x+2\right)=0\\ \left[{}\begin{matrix}x-3=0\\4x+2=0\end{matrix}\right.=>x=3;x=-\dfrac{1}{2}\)
\(\left(3x-1\right)\left(x-3\right)-9+x^2=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(3x-1\right)\left(x-3\right)-9+x^2=0\)
\(\Leftrightarrow3x^2-9x-x+3-9+x^3=0\)
\(\Leftrightarrow4x^2-10x-6=0\)
\(\Leftrightarrow4x^2-12+2x-6=0\)
\(\Leftrightarrow4x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{3;\dfrac{-1}{2}\right\}\).
(3x - 1)(x - 3) - 9 + x^2 = 0
<=> (3x - 1)(x - 3)-(3 - x)(3 + x)=0
<=> (3x - 1)(x - 3)+(x - 3)(3+x)=0
<=> (x - 3)(3x - 1 + 3 + x) = 0
=> x - 3 = 0 hoặc 4x + 2 = 0
<=> x = 0 + 3 hoặc 4x = - 2
<=> x = 3 hoặc x = - 1/2
Vậy:...
