Theo đề bài, ta có:
\(\dfrac{a+b}{b+b}=2.\dfrac{a}{b},\Rightarrow\dfrac{a+b}{2b}=\dfrac{2a}{b}\), do đó \(b\left(a+b\right)=4ab\) hay \(ab+b^2=4ab\) \(\Leftrightarrow\)\(3ab=b^2\Rightarrow3a=b\). Ta có \(\dfrac{a}{b}=\dfrac{1}{3}\).
Vậy phân số cần tìm là: \(\dfrac{1}{3}\).
Theo đề bài ta có:
\(2\dfrac{a}{b}=\dfrac{a+b}{b+b}\)
\(\Rightarrow\dfrac{2a}{b}=\dfrac{a+b}{2b}\)
\(\Rightarrow\dfrac{4a}{2b}=\dfrac{a+b}{2b}\)
\(\Rightarrow a+b=4a\)
\(\Rightarrow b=3a\)
\(\Rightarrow a\in\left\{1;2;3;....\right\}\)
\(\Rightarrow b\in\left\{3;6;9;....\right\}\)
\(\left(a;b\right)=1\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{1}{3}\)
