Cho ${{\log }_{9}}5=a,{{\log }_{4}}7=b,{{\log }_{2}}3=c$. Biết...

Ta có
$\begin{aligned}
& {{\log }_{24}}175={{\log }_{{{2}^{3}}.3}}{{5}^{2}}.7={{\log }_{{{2}^{3}}.3}}{{5}^{2}}+{{\log }_{{{2}^{3}}.3}}7 \\
& =\dfrac{2}{{{\log }_{5}}{{2}^{3}}.3}+\dfrac{1}{{{\log }_{7}}{{2}^{3}}.3}=\dfrac{2}{3.{{\log }_{5}}2+{{\log }_{5}}3}+\dfrac{1}{3{{\log }_{7}}2+{{\log }_{7}}3} \\
\end{aligned}$

Theo giả thiết ta có:
$\left\{ \begin{aligned}
& {{\log }_{9}}5=a\Rightarrow {{\log }_{3}}5=2a \\
& {{\log }_{4}}7=b\Rightarrow {{\log }_{2}}7=2b \\
& {{\log }_{2}}3=c \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\log }_{7}}3=\dfrac{c}{2b} \\
& {{\log }_{5}}3=\dfrac{1}{2a} \\
& {{\log }_{5}}2=\dfrac{1}{2ac} \\
\end{aligned} \right.$.

Suy ra:
${{\log }_{24}}175=\dfrac{2}{\dfrac{3}{2ac}+\dfrac{1}{2a}}+\dfrac{1}{\dfrac{3}{2b}+\dfrac{c}{2b}}=\dfrac{2}{\dfrac{3+c}{2ac}}+\dfrac{1}{\dfrac{3+c}{2b}}=\dfrac{4ac}{3+c}+\dfrac{2b}{3+c}=\dfrac{4ac+2b}{c+3}$.
Vậy ta có: $\left\{ \begin{aligned}
& m=2 \\
& n=4 \\
& p=1 \\
& q=3 \\
\end{aligned} \right.\Rightarrow mnpq=24$.