Ta có:
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)
\(=\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(=\frac{7}{x}+\frac{1}{5}-\frac{1}{45}\)
\(=\frac{7}{x}+\frac{9}{45}-\frac{1}{45}=\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\)
\(\Rightarrow x=15\)
Vậy x=15