Giải phương trình $\log_{3}\left( {2x-1} \right) = 2$ , ta có nghiệm là:

$x=3+\sqrt{2}$               

$x=\dfrac{-11}{4}$         

$x=3-\sqrt{2}$                 

$x=\dfrac{11}{4}$

$\left\{ {\dfrac{{1 + \sqrt 2 }}{2}} \right\}$

$\left\{ {2;41} \right\}$

$\left\{ {1 - \sqrt 2 ;1 + \sqrt 2 } \right\}$     

$\left\{ {1 + \sqrt 2 } \right\}$ 

$x = 1$          

$x = \sqrt[8]{{{3^5}}} - 2$    

$x = \sqrt[4]{{{3^5}}} - 2$

$x = \sqrt[4]{3} - 2.$

$x = 63$

$x = 65$

$x = 82$ 

$x = 80$

$S = \left\{ { - 3;3} \right\}$.

$S = \left\{ {\sqrt {10} } \right\}$.

$S = \left\{ 3 \right\}$.

$S = \left\{ { - \sqrt {10} ;\sqrt {10} } \right\}$.

$\dfrac{{ - 1 + \sqrt 5 }}{2}.$

$\dfrac{{ - 1 - \sqrt 5 }}{2}.$

$\dfrac{{1 + \sqrt 5 }}{2}.$

$\dfrac{1}{2}.$