Giải:
Ta có: \(\frac{a}{b}=\frac{9}{4}\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)
\(\frac{b}{c}=\frac{5}{3}\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{45}=\frac{b}{20}=\frac{c}{12}\)
Đặt \(\frac{a}{45}=\frac{b}{20}=\frac{c}{12}=k\Rightarrow\left\{\begin{matrix}a=45k\\b=20k\\c=12k\end{matrix}\right.\)
\(\frac{a-b}{b-c}=\frac{45k-20k}{20k-12k}=\frac{\left(45-20\right)k}{\left(20-12\right)k}=\frac{25}{8}\)
Vậy \(\frac{a-b}{b-c}=\frac{25}{8}\)
Ta có \(\frac{a}{b}=\frac{9}{4}\)\(\Rightarrow\frac{a}{9}=\frac{b}{4}\Rightarrow\frac{a}{45}=\frac{b}{20}\)
\(\frac{b}{c}=\frac{5}{3}\Rightarrow\frac{b}{5}=\frac{c}{3}\Rightarrow\frac{b}{20}=\frac{c}{15}\)
Khi đó ta có \(\frac{a}{45}=\frac{b}{20}=\frac{c}{15}\)
Áp dụng tính chất dãy tỉ số bừng nhau ta có
\(\frac{a}{45}=\frac{b}{20}=\frac{a-b}{15}\left(1\right)\)
\(\frac{b}{20}=\frac{c}{15}=\frac{b-c}{5}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\) ta có
\(\frac{a-b}{15}=\frac{b-c}{5}\Rightarrow\frac{a-b}{b-c}=\frac{15}{5}=3\)
