a,\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
b,\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008
