\(\frac{8}{11}+\frac{a}{b}=\frac{8}{11}\)
\(\Rightarrow\frac{a}{b}=0\)
\(\Rightarrow a=0;b\in Z;b\ne0\)
\(\frac{8}{11}+\frac{a}{b}=\frac{8}{11}\cdot\frac{a}{b}\)
=> \(\frac{8b+11a}{11b}=\frac{8a}{11b}\)
=> 8b + 11a = 8a
=> 8b = 8a - 11a
=> 8b - 8a + 11a = 0
=> 8b + 3a = 0
=> 8b = -3a
=> \(\frac{a}{8}=\frac{b}{-3}\)=> \(\frac{a}{b}=\frac{8}{-3}\)hay \(\frac{a}{b}=\frac{-8}{3}\)
Vậy \(\frac{a}{b}=\frac{-8}{3}\)