Question

tìm x:\(\frac{32}{8.11}+\frac{32}{11.14}+\frac{32}{14.17}+...+\frac{32}{197.200}-x=\frac{1}{2}\)

Answer 1

\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)

x=0.78