Question

tìm x để \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{10300}=\frac{1}{x}\)

Answer 1

\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{10300}=\frac{1}{x}\)

=> \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{100\cdot103}=\frac{1}{x}\)

=> \(\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{100\cdot103}\right)=\frac{1}{x}\)

=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{1}{x}\)

=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{103}\right)=\frac{1}{x}\)

=> \(\frac{101}{618}=\frac{1}{x}\)

=> \(101x=618\)

=> \(x=\frac{618}{101}\)

Vậy : ...

Answer 2

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{10300}=\frac{1}{x}\)

\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{100.103}=3.\frac{1}{x}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}=3.\frac{1}{x}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{103}=3.\frac{1}{x}\)

\(\Rightarrow\frac{1}{x}.3=\frac{101}{206}\)

\(\Rightarrow\frac{1}{x}=\frac{101}{618}\)

\(\Rightarrow x=\frac{618}{101}\)