sửa lại đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(S=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(S=1-\frac{1}{17}=\frac{16}{17}\)
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{14}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\)
\(\Rightarrow S=1-\frac{1}{17}\)
\(\Rightarrow S=\frac{16}{17}\)
Vậy S = \(\frac{16}{17}\)
S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
S=\(3.\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-.....+\frac{1}{14}-\frac{1}{17}\right)\)
S=\(3.\frac{1}{3}.\left(1-\frac{1}{17}\right)\)
S=\(3.\frac{1}{3}.\frac{16}{17}\)
S=\(1.\frac{16}{17}\)
S=\(\frac{16}{17}\)
Giải:
Ta có: S=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(\Leftrightarrow S=1-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-....-\left(\frac{1}{14}-\frac{1}{14}\right)-\frac{1}{17}\)
\(\Leftrightarrow S=1-0-0-...-0-\frac{1}{17}\)
\(\Leftrightarrow S=1-\frac{1}{17}=\frac{16}{17}\)
Vậy: \(S=\frac{16}{17}\)
