`@Fun` 

`6.4^x - 13.6^x + 6.9^x = 0` 

`<=> 6. (4/4)^x - 13. (6/4)^x + 6 . (9/4)^x = 0` 

`<=> 6 - 13 . (3/2)^x + 6 . (3/2)^(2x) = 0` 

Đặt : `(3/2)^x = t , t> 0` 

`<=> 6 - 13t + 6t^2 = 0` 

`<=>` \(\left[ \begin{array}{l}t=\dfrac{3}{2}\\t=\dfrac{2}{3}\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}(\dfrac{3}{2})^x=\dfrac{3}{2}\\(\dfrac{3}{2})^x=\dfrac{2}{3}\end{array} \right.\)

`<=> ` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

Vậy pt có `n_o : x  = 1 ; x = -1`