\(\frac{9+\frac{9}{29}-\frac{1}{3}+\frac{1}{5}-\frac{18}{100}}{3+\frac{3}{29}-\frac{1}{9}+\frac{1}{15}-\frac{6}{100}}\)=\(\frac{3.\left(3+\frac{3}{29}-\frac{1}{9}+\frac{1}{15}-\frac{6}{100}\right)}{3+\frac{3}{29}-\frac{1}{9}+\frac{1}{15}-\frac{6}{100}}\)
=3