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\(A< \frac{1}{1\cdot3}+\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+..........+\frac{1}{2011\cdot2013}\)
\(\frac{1}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+.....+\frac{1}{2010}-\frac{1}{2013}\right)\)
\(\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}\cdot\frac{2012}{2013}\)
theo mình là vậy thôi chứ ko chắc chắn đouo
bạn nhok ma kết làm gần đúng nhưng vẫn sai nhé
Đặt biểu thức là A
\(A=\frac{1}{9}\left(\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{671^2}\right)< \frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{671.672}\right)\)
\(\Rightarrow A< \frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{671}-\frac{1}{672}\right)\)
\(\Rightarrow A< \frac{1}{9}\left(1-\frac{1}{672}\right)=\frac{1}{9}.\frac{671}{672}< \frac{1}{5}.1=\frac{1}{5}\)
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}:\frac{1}{\frac{7}{2}}=\frac{2}{7}:\frac{2}{7}=1\)
\(2.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2013}+\frac{1}{2015}+\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=2.\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)-1\)
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
= \(\frac{5}{9}\) .(\(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\) )
= \(\frac{5}{9}\) . 1 = \(\frac{5}{9}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{14}{15}\)
\(=\frac{7}{15}\)
Sửa đề chút nhé:
\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).0\)
\(=0\)
Ý b tham khảo bài bạn nguyen thi thuy linh nhé
\(P=\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{2013}\right)\)
\(P=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{2011}{2013}\)
\(P=\frac{3}{2013}\)