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Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
=>\(S=\frac{4}{9}-\frac{1}{5}\)
=>\(S=\frac{11}{45}\)
A =
A = \(1-\frac{1}{2018}\)
A = \(\frac{2017}{2018}\)
Có :
2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
2.B = \(1-\frac{1}{2017}\)
2.B = \(\frac{2016}{2017}\)
B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)
Có :
3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)
3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)
3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
1 \(-\)\(\frac{1}{3.5}\)\(-\)\(\frac{1}{5.7}\)\(-\)\(\frac{1}{7.9}\)\(-\)..... \(-\)\(\frac{1}{53.55}\)\(-\)\(\frac{1}{55.57}\)
= 1 \(-\)( \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) + \(\frac{1}{7.9}\) + ..... + \(\frac{1}{53.55}\) + \(\frac{1}{55.57}\) )
= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)+ \(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)+ \(\frac{1}{7}\)\(-\)\(\frac{1}{9}\)+....+ \(\frac{1}{53}\)\(-\)\(\frac{1}{55}\)+ \(\frac{1}{55}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)
= 1 \(-\)( \(\frac{1}{3}\)\(-\)\(\frac{1}{57}\)) . \(\frac{1}{2}\)
= 1 \(-\) \(\frac{6}{19}\). \(\frac{1}{2}\)= 1 \(-\)\(\frac{3}{19}\)= \(\frac{16}{19}\)
\(1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)
đặt \(A=1-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-...-\frac{1}{53.55}-\frac{1}{55.57}\)
\(A=1-\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)
đặt \(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{53.55}+\frac{1}{55.57}\)
\(2B=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{53.55}+\frac{1}{55.57}\right)\)
\(2B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{53.55}+\frac{2}{55.57}\)
\(2B=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{55-53}{53.55}+\frac{57-55}{55.57}\)
\(2B=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+\frac{9}{7.9}-\frac{7}{7.9}+...+\frac{55}{53.55}-\frac{53}{53.55}+\frac{57}{55.57}-\frac{55}{55.57}\)
\(2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(2B=\frac{1}{3}-\frac{1}{57}\)
\(2B=\frac{54}{171}\)
\(\Rightarrow B=\frac{54}{171}:2\)
\(\Rightarrow B=\frac{9}{57}\)
mà \(A=1-B\)
\(\Rightarrow A=1-\frac{9}{57}\)
\(\Rightarrow A=\frac{48}{57}\)
chúc bạn học giỏi ^^
\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{2017}\right)\)
\(=2.\frac{2014}{6051}\)
\(=\frac{4028}{6051}\)
\(\Rightarrow BT>\frac{1}{6}\)