✓3х-1=1-3х.памогите решить

Решите задачу:

$\displaystyle \sqrt{3x-1}=1-3x\\\\\sqrt{(3x-1)^2}=(1-3x)^2\\\\3x-1-(3x-1)^2=0\\\\(3x-1)(1-(3x-1))=0\\\\(3x-1)(1-3x+1)=0\\\\(3x-1)(2-3x)=0\\\\ \left[\begin{array}{ccc}\displaystyle 3x-1=0\\2-3x=0\end{array}\right\rightarrow \left[\begin{array}{ccc}\displaystyle 3x=1\\3x=2\end{array}\right\rightarrow {\left[\begin{array}{ccc}\displaystyle x=\frac{1}3\\\\\displaystyle x=\frac{2}3\end{array}$

$\displaystyle \text{ODZ}:\\\\\quad 1-3x \geq 0\\\\3x \leq 1\\\\x \leq \frac{1}3$

$\boxed{\displaystyle x=\frac{1}3}$