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

$x=3+\sqrt{2}$               

$x=\dfrac{-11}{4}$         

$x=3-\sqrt{2}$                 

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

\(\left\{ {\dfrac{{1 + \sqrt 2 }}{2}} \right\}\)

\(\left\{ {2;41} \right\}\)

\(\left\{ {1 - \sqrt 2 ;1 + \sqrt 2 } \right\}\)     

\(\left\{ {1 + \sqrt 2 } \right\}\) 

\(x = 1\)          

\(x = \sqrt[8]{{{3^5}}} - 2\)    

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

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

$x = 63$

$x = 65$

$x = 82$ 

$x = 80$

\(S = \left\{ { - 3;3} \right\}\).

\(S = \left\{ {\sqrt {10} } \right\}\).

\(S = \left\{ 3 \right\}\).

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

\(\dfrac{{ - 1 + \sqrt 5 }}{2}.\)

\(\dfrac{{ - 1 - \sqrt 5 }}{2}.\)

\(\dfrac{{1 + \sqrt 5 }}{2}.\)

\(\dfrac{1}{2}.\)