Đáp án:
$\begin{array}{l}
\log _5^2\dfrac{x}{5} + \left( {m + 1} \right){\log _5}5x + 6m - 22 = 0\\
\Rightarrow {\left( {{{\log }_5}x - 1} \right)^2} + \left( {m + 1} \right)\left( {1 + {{\log }_5}x} \right) + 6m - 22 = 0\\
Dat:{\log _5}x = t\\
Do:x \in \left[ {\dfrac{1}{5};{5^5}} \right]\\
\Rightarrow t \in \left[ { - 1;5} \right]\\
pt \Rightarrow {\left( {t - 1} \right)^2} + \left( {m + 1} \right).\left( {1 + t} \right) + 6m - 22 = 0\\
\Rightarrow {t^2} + \left( {m - 1} \right)t + 7m - 20 = 0\\
\Rightarrow {t^2} - t - 20 + \left( {t + 7} \right).m = 0\\
\Rightarrow \dfrac{{ - {t^2} + t + 20}}{{t + 7}} = m\left( {do:t \ne - 7} \right)\\
f\left( t \right) = \dfrac{{ - {t^2} + t + 20}}{{t + 7}}\\
\Rightarrow f'\left( t \right) = \dfrac{{\left( { - 2t + 1} \right)\left( {t + 7} \right) + {t^2} - t - 20}}{{{{\left( {t + 7} \right)}^2}}}\\
= \dfrac{{ - {t^2} - 14t - 13}}{{{{\left( {t + 7} \right)}^2}}} < 0\left( {khi:t \in \left[ { - 1;5} \right]} \right)
\end{array}$
=> f(t) nghịch biến trên [-1;5]
$\begin{array}{l}
\log _5^2\dfrac{x}{5} + \left( {m + 1} \right){\log _5}5x + 6m - 22 = 0\\
\Rightarrow {\left( {{{\log }_5}x - 1} \right)^2} + \left( {m + 1} \right)\left( {1 + {{\log }_5}x} \right) + 6m - 22 = 0\\
Dat:{\log _5}x = t\\
Do:x \in \left[ {\dfrac{1}{5};{5^5}} \right]\\
\Rightarrow t \in \left[ { - 1;5} \right]\\
pt \Rightarrow {\left( {t - 1} \right)^2} + \left( {m + 1} \right).\left( {1 + t} \right) + 6m - 22 = 0\\
\Rightarrow {t^2} + \left( {m - 1} \right)t + 7m - 20 = 0\\
\Rightarrow {t^2} - t - 20 + \left( {t + 7} \right).m = 0\\
\Rightarrow \dfrac{{ - {t^2} + t + 20}}{{t + 7}} = m\left( {do:t \ne - 7} \right)\\
\Rightarrow f\left( 5 \right) \le m \le f\left( { - 1} \right)\\
\Rightarrow 0 \le m \le 3
\end{array}$
