DK: x≠ 1/3,-1/3
pt<=> \(\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
=> 12=(1-3x)2-(1+3x)2=(1-3x-1-3x)(1-3x+1+3x)
=(-6x).2=-12x
=> x=-1
\(\text{Đ}KX\text{Đ}:\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.=>\left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right)=>\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)\(< =>12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(< =>12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
<=>12x+12=0
<=>12x=-12
<=>x=-1(nhận)
\(S=\left\{-1\right\}\)
