Giải:
Ta có: \(\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)
\(\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)
Đặt \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow\left[\begin{matrix}a=45k\\b=20k\\c=12k\end{matrix}\right.\)
Lại có: \(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{25k}{8k}=\frac{25}{8}\)
Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)
Theo bài ra:
\(\dfrac{a}{b}=\dfrac{9}{4}\Rightarrow a=\dfrac{9}{4}.b\)
\(\dfrac{b}{c}=\dfrac{5}{3}\Rightarrow c=b:\dfrac{5}{3}\)
Thay \(a=\dfrac{9}{4b};c=b:\dfrac{5}{3}\) vào \(\dfrac{a-b}{b-c}\), ta có:
\(\dfrac{\dfrac{9b}{4}-b}{b-\dfrac{3b}{5}}=\dfrac{\dfrac{9b}{4}-\dfrac{4b}{4}}{\dfrac{5b}{5}-\dfrac{3b}{5}}=\dfrac{5b}{4}:\dfrac{2b}{5}=\dfrac{5b}{4}.\dfrac{5}{2b}=\dfrac{25}{8}\)
Vậy: \(\dfrac{a-b}{b-c}=\dfrac{25}{8}\)
