$\sqrt{\dfrac{9}{25}x^2-\dfrac{6}{5}x+1}=9$
$↔\dfrac{9}{25}x^2-\dfrac{6}{5}x+1=81$
$↔\Big(\dfrac{3}{5}x\Big)^2-2.\dfrac{3}{5}x.1+1^2=81$
$↔\Big(\dfrac{3}{5}x-1\Big)^2=81$
\(→\left[ \begin{array}{l}\dfrac{3}{5}x-1=81\\\dfrac{3}{5}x-1=-81\end{array} \right.\)
\(→\left[ \begin{array}{l}\dfrac{3}{5}x=81\\\dfrac{3}{5}x=-80 \end{array} \right.\)
\(→\left[ \begin{array}{l}x=81:\dfrac{3}{5}\\x=-80:\dfrac{3}{5}\end{array} \right.\)
\(→\left[ \begin{array}{l}x=135\\x=-\dfrac{400}{3}\end{array} \right.\)
Vậy $x=\Big\{135;-\dfrac{400}{3}\Big\}$
