Ta có:\(\frac{a}{b-a}=8.\frac{a}{b}\)
\(\Rightarrow\frac{a}{b-a}=\frac{8.a}{b}\)
hay\(\frac{a.b}{\left(b-a\right).b}=\frac{\left(b-a\right).8.a}{\left(b-a\right).b}\)
\(\Rightarrow a.b=\left(b-a\right).8.a\)
\(\Rightarrow a.b=8.a.b-8.a^2\)
\(\Rightarrow a.b\div a=\left(8.a.b-8.a^2\right)\div a\)
\(\Rightarrow b=8.b-8.a\)
\(\Rightarrow8.b-b=8.a\)
\(\Rightarrow7.b=8.a\)
\(\Rightarrow\hept{\begin{cases}a=7\\b=8\end{cases}}\)
Vậy\(\hept{\begin{cases}a=7\\b=8\end{cases}}\)