Question

Tìm x ; 

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\) \(\frac{29}{45}\)

Answer 1

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)

\(\Rightarrow x=\frac{7}{\frac{21}{45}}=15\)

Vậy \(x=15\).