Biết $a={\log }_{27}5$, $b={\log }_{8}7$, $c={\log }_{2}3$.

a) Sai: Ta có: $c={\log }_{2}3<{\log }_{2}4=2$.
b) Đúng: Ta có: $a.c=c.a={\log }_2(3){\log }_{27}(5)={\log }_2(3)\cdot \frac{1}{3}{\log }_3(5)=\frac{1}{3}{\log }_{2}5$.
c) Đúng: Ta có: $\frac{a.c}{b}=\frac{\frac{1}{3}{\log }_{2}5}{{\log }_{8}7}=\frac{1}{3}{\log }_2(5).{\log }_7(8)=\frac{1}{3}{\log }_2(5).3.{\log }_7(2)={\log }_{7}5$.
d) Đúng: Ta có: $\{\begin{array}{l}a={\log }_{27}(5)=\frac{1}{3}{\log }_{3}5\\ b={\log }_{8}7=\frac{1}{3}{\log }_{2}7\end{array}\iff \{\begin{array}{l}{\log }_{3}5=3a\\ {\log }_{2}7=3b\end{array}$
Mà: ${\log }_{12}35=\frac{{\log }_2(7.5)}{{\log }_2\left(3.2^2\right)}=\frac{{\log }_{2}7+{\log }_{2}5}{{\log }_{2}3+2}=\frac{{\log }_{2}7+{\log }_{2}3.{\log }_{3}5}{{\log }_{2}3+2}=\frac{3b+c.3a}{c+2}=\frac{3(b+ac)}{c+2}$.
a) Sai: Ta có: $c={\log }_{2}3<{\log }_{2}4=2$.
b) Đúng: Ta có: $a.c=c.a={\log }_2(3){\log }_{27}(5)={\log }_2(3)\cdot \frac{1}{3}{\log }_3(5)=\frac{1}{3}{\log }_{2}5$.
c) Đúng: Ta có: $\frac{a.c}{b}=\frac{\frac{1}{3}{\log }_{2}5}{{\log }_{8}7}=\frac{1}{3}{\log }_2(5).{\log }_7(8)=\frac{1}{3}{\log }_2(5).3.{\log }_7(2)={\log }_{7}5$.
d) Đúng: Ta có: $\{\begin{array}{l}a={\log }_{27}(5)=\frac{1}{3}{\log }_{3}5\\ b={\log }_{8}7=\frac{1}{3}{\log }_{2}7\end{array}\iff \{\begin{array}{l}{\log }_{3}5=3a\\ {\log }_{2}7=3b\end{array}$
Mà: ${\log }_{12}35=\frac{{\log }_2(7.5)}{{\log }_2\left(3.2^2\right)}=\frac{{\log }_{2}7+{\log }_{2}5}{{\log }_{2}3+2}=\frac{{\log }_{2}7+{\log }_{2}3.{\log }_{3}5}{{\log }_{2}3+2}=\frac{3b+c.3a}{c+2}=\frac{3(b+ac)}{c+2}$.