Đặt \(\frac{a}{b}=k\)
Theo bài ra ta có:
\(k=\left(\frac{7}{18}+\frac{11}{8}+k\right)\div3\)
\(\Rightarrow3k=\frac{127}{72}+k\)
\(\Rightarrow2k=\frac{127}{72}\)
\(\Leftrightarrow k=\frac{127}{144}\)
Vậy, \(\frac{a}{b}=\frac{127}{144}\)
Ta có: \(\frac{a}{b}=\left(\frac{7}{18}+\frac{11}{8}+\frac{a}{b}\right):3\)
\(\Leftrightarrow3.\frac{a}{b}=\frac{7}{18}+\frac{11}{8}+\frac{a}{b}\)\(\Leftrightarrow3.\frac{a}{b}-\frac{a}{b}=\frac{7}{18}+\frac{11}{8}\)
\(\Leftrightarrow2.\frac{a}{b}=\frac{127}{72}\)\(\Leftrightarrow\frac{a}{b}=\frac{127}{144}\)
Vậy \(\frac{a}{b}=\frac{127}{144}\)