\(\sqrt{9x^2-6x+1}=-x\\ < =>\sqrt{\left(3x-1\right)^2}=-x\\ < =>\left|3x-1\right|=-x\\ \)
Ta có : \(\left\{{}\begin{matrix}3x-1\ge0< =>3x\ge1< =>x\ge\dfrac{1}{3}\\3x-1< 0< =>3x< 1< =>x< \dfrac{1}{3}\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}3x-1=-x\\-\left(3x-1\right)=-x\end{matrix}\right.\\ < =>\left[{}\begin{matrix}3x+x=1\\-3x+1=-x\end{matrix}\right.\\ < =>\left[{}\begin{matrix}4x=1\\-3x+x=-1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(ktm\right)\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\)
=>|3x-1|=-x
TH1: x>=1/3
Pt sẽ là 3x-1=-x
=>4x=1
=>x=1/4(loại)
TH2: x<1/3
Pt sẽ là 1-3x=-x
=>3x-1=x
=>2x=1
=>x=1/2(loại)
