Cho $a={\log }_{2}5$, $b={\log }_{3}5$.

Đáp án: a) Đúng, b) Sai, c) Sai, d) Đúng.
a) Đúng: Ta có $b={\log }_{3}5>{\log }_{3}3=1$
b) Sai: Ta có ${\log }_{2}3=\frac{{\log }_{5}3}{{\log }_{5}2}=\frac{1}{{\log }_{3}5}.{\log }_{2}5=\frac{a}{b}$.
c) Sai: ${\log }_{12}5=\frac{1}{{\log }_{5}12}=\frac{1}{{\log }_5\left(2^2.3\right)}=\frac{1}{{\log }_5{2}^2+{\log }_{5}3}=\frac{1}{\frac{2}{{\log }_{2}5}+\frac{1}{{\log }_{3}5}}=\frac{1}{\frac{2}{a}+\frac{1}{b}}=\frac{1}{\frac{a+2b}{ab}}=\frac{ab}{a+2b}$.
d) Đúng: ${\log }_{24}250={\log }_{2^3.3}\left(2.5^3\right)={\log }_{2^3.3}2+{\log }_{2^3.3}{5}^3=\frac{1}{{\log }_2\left(2^3.3\right)}+\frac{3}{{\log }_5\left(2^3.3\right)}$
$$=\frac{1}{3+{\log }_{2}3}+\frac{3}{3{\log }_{5}2+{\log }_{5}3}=\frac{1}{3+\frac{{\log }_{2}5}{{\log }_{3}5}}+\frac{3}{\frac{3}{{\log }_{2}5}+\frac{1}{{\log }_{3}5}}$$
$$=\frac{1}{3+\frac{a}{b}}+\frac{3}{\frac{3}{a}+\frac{1}{b}}=\frac{1}{\frac{3b+a}{b}}+\frac{3}{\frac{3b+a}{ab}}=\frac{b}{3b+a}+\frac{3ab}{3b+a}=\frac{3ab+b}{a+3b}.$$
Vậy ta có $\{\begin{array}{l}m=3\\ n=1\\ p=1\\ q=3\end{array}\Rightarrow mnpq=9$.
a) Đúng: Ta có $b={\log }_{3}5>{\log }_{3}3=1$
b) Sai: Ta có ${\log }_{2}3=\frac{{\log }_{5}3}{{\log }_{5}2}=\frac{1}{{\log }_{3}5}.{\log }_{2}5=\frac{a}{b}$.
c) Sai: ${\log }_{12}5=\frac{1}{{\log }_{5}12}=\frac{1}{{\log }_5\left(2^2.3\right)}=\frac{1}{{\log }_5{2}^2+{\log }_{5}3}=\frac{1}{\frac{2}{{\log }_{2}5}+\frac{1}{{\log }_{3}5}}=\frac{1}{\frac{2}{a}+\frac{1}{b}}=\frac{1}{\frac{a+2b}{ab}}=\frac{ab}{a+2b}$.
d) Đúng: ${\log }_{24}250={\log }_{2^3.3}\left(2.5^3\right)={\log }_{2^3.3}2+{\log }_{2^3.3}{5}^3=\frac{1}{{\log }_2\left(2^3.3\right)}+\frac{3}{{\log }_5\left(2^3.3\right)}=\frac{1}{3+{\log }_{2}3}+\frac{3}{3{\log }_{5}2+{\log }_{5}3}=\frac{1}{3+\frac{{\log }_{2}5}{{\log }_{3}5}}+\frac{3}{\frac{3}{{\log }_{2}5}+\frac{1}{{\log }_{3}5}}=\frac{1}{3+\frac{a}{b}}+\frac{3}{\frac{3}{a}+\frac{1}{b}}=\frac{1}{\frac{3b+a}{b}}+\frac{3}{\frac{3b+a}{ab}}=\frac{b}{3b+a}+\frac{3ab}{3b+a}=\frac{3ab+b}{a+3b}$. Vậy ta có $\{\begin{array}{l}m=3\\ n=1\\ p=1\\ q=3\end{array}\Rightarrow mnpq=9$.