Cho phương trình $\left(2 \log _{3}^{2} x-\log _{3} x-1\right)...

Điều kiện .
$\left( 2\log _{3}^{2}x-{{\log }_{3}}x-1 \right)\sqrt{{{5}^{x}}-m}=0\Leftrightarrow \left\{ \begin{aligned}
& x>0,\ {{5}^{x}}-m\ge 0 \\
& \left[ \begin{aligned}
& {{5}^{x}}-m=0 \\
& 2\log _{3}^{2}x-{{\log }_{3}}x-1=0 \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x>0,\ {{5}^{x}}-m\ge 0 \\
& \left[ \begin{aligned}
& {{5}^{x}}-m=0 \\
& {{\log }_{3}}x=\dfrac{-1}{2} \\
& {{\log }_{3}}x=1 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0,{{5}^{x}}-m\ge 0 \\
& \left[ \begin{aligned}
& x={{\log }_{5}}m \\
& x={{3}^{\dfrac{1}{\sqrt{2}}}} \\
& x=3 \\
\end{aligned} \right. \\
\end{aligned} \right.m=1\Rightarrow x={{\log }_{2}}1=0\left(2 \log _{3}^{2} x-\log _{3} x-1\right) \sqrt{5^{x}-m}=0\left[ \begin{aligned}
& x={{3}^{\dfrac{1}{\sqrt{2}}}} \\
& x=3 \\
\end{aligned} \right.m>1\Rightarrow x={{\log }_{5}}m\dfrac{1}{\sqrt{3}}\le {{\log }_{5}}m<3\Leftrightarrow {{5}^{\dfrac{1}{\sqrt{3}}}}\le m<{{5}^{3}}\Leftrightarrow 2,53\le m<125$