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\(\frac{1}{3}-\left(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+\frac{1}{180}+\frac{1}{270}+\frac{1}{378}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{3}-\frac{1}{3}+\frac{1}{21}=\frac{1}{21}\)
\(B=1.\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{3}.\frac{1}{4}+...+\frac{1}{8}.\frac{1}{9}.\frac{1}{10}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)
\(B=\frac{1}{2}.\frac{22}{45}\)
\(B=\frac{11}{45}\)
\(B=1.\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{3}.\frac{1}{4}+....+\frac{1}{8}.\frac{1}{9}.\frac{1}{10}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=\frac{1}{2}.\left(1-\frac{1}{10}\right)\)
\(B=\frac{1}{2}.\frac{9}{10}\)
\(B=\frac{9}{20}\)
A=\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\) =\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\) =\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{2}.\frac{10}{33}=\frac{5}{33}\)
\(A=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(A=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{11}\right)\)
\(A=\frac{1}{9}.\frac{10}{11}=\frac{10}{99}\)
Đặt \(S=\frac{A}{B}\)
Biến đổi B
\(B=\frac{108}{1}+\frac{107}{2}+...+\frac{1}{108}\)
\(=\left(\frac{108}{1}+1\right)+\left(\frac{107}{2}+1\right)+...+\left(\frac{1}{108}+1\right)-108\)
\(=109+\frac{109}{2}+...+\frac{109}{108}-108\)
\(=109+\frac{109}{2}+...+\frac{109}{108}+\frac{109}{109}-109\)
\(=109.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}\right)\)
\(\Rightarrow s=\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}}{109.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{109}\right)}=\frac{1}{109}\)
KO hiểu em hỏi nhé
Em ko cần đặt \(S=\frac{A}{B}\)cũng được nhé tại vì anh có thói quen đặt
=\(\frac{85}{504}\)
\(\frac{85}{504}\)