Cho các số thực $a,b$ thỏa mãn $a>b>1$ và $\dfrac{1}{{{\log...

Ta có:
$\begin{aligned}
& \dfrac{1}{{{\log }_{b}}a}+\dfrac{1}{{{\log }_{a}}b}=\sqrt{2024}\Leftrightarrow \dfrac{1}{{{\log }_{b}}a}+{{\log }_{b}}a=2\sqrt{506}\Leftrightarrow {{\left( {{\log }_{b}}a \right)}^{2}}-2\sqrt{506}{{\log }_{b}}a+1=0 \\
& \begin{matrix}
\begin{matrix}
\begin{matrix}
\begin{matrix}
\begin{matrix}
{} & {} & {} & {} \\
\end{matrix} & {} & {} & {} \\
\end{matrix} & {} & {} & {} \\
\end{matrix} & {} & {} & {} \\
\end{matrix} & {} & {} & \Leftrightarrow \left[ \begin{aligned}
& {{\log }_{b}}a=\sqrt{506}+\sqrt{505} \\
& {{\log }_{b}}a=\sqrt{506}-\sqrt{505} \\
\end{aligned} \right. \\
\end{matrix} \\
\end{aligned}$.
Ta có $P=\dfrac{1}{{{\log }_{ab}}b}-\dfrac{1}{{{\log }_{ab}}a}={{\log }_{b}}ab-{{\log }_{a}}ab=1+{{\log }_{b}}a-1-{{\log }_{a}}b$.
+) Với ${{\log }_{b}}a=\sqrt{506}-\sqrt{505}$. Suy ra:
${{\log }_{a}}b=\dfrac{1}{\sqrt{506}-\sqrt{505}}\Rightarrow P=\dfrac{-1}{\sqrt{506}-\sqrt{505}}+\dfrac{1}{\sqrt{506}+\sqrt{505}}=-2\sqrt{505}$ (loại).
+) Với ${{\log }_{b}}a=\sqrt{506}+\sqrt{505}$. Suy ra:
${{\log }_{a}}b=\dfrac{1}{\sqrt{506}+\sqrt{505}}\Rightarrow P=\sqrt{506}+\sqrt{505}-\dfrac{1}{\sqrt{506}+\sqrt{505}}=\dfrac{1}{\sqrt{506}-\sqrt{505}}-\dfrac{1}{\sqrt{506}+\sqrt{505}}=2\sqrt{505}$ (thỏa mãn).
Vậy $P=\dfrac{1}{{{\log }_{ab}}b}-\dfrac{1}{{{\log }_{ab}}a}=2\sqrt{505}=\sqrt{2020}$.