Question

[Hoàng Lê Ngọc Mai]

Tìm x

a. x/2008-1/10-1/15-1./21-...-1/120=5/8

b. 7/x+4/5.9+4/9.13+4/13.17+.....+4/41.45=29/45

Answer 1

[Yen Nhi]

\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{5}-\frac{1}{16}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2.\frac{3}{16}]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}=1\)

\(\Rightarrow x=2008\)

Answer 2

[Yen Nhi]

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

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

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

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

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

\(\Rightarrow x=15\)