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\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)
\(B=1-\frac{1}{59}\)
\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)
Vậy B = \(\frac{58}{59}\)
Lưu ý: Dấu "." là dấu nhân
\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)
\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)
\(B=1.\left(1-\frac{1}{59}\right)\)
\(B=1.\frac{58}{59}\)
\(B=\frac{58}{59}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\frac{2016}{2017}\)
\(S=\frac{1008}{2017}< \frac{1}{2}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(2S=1-\frac{1}{2017}< 1\)
=> 2S < 1
=> S < \(\frac{1}{2}\)(đpcm)
\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)
= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)
= \(1.486077098\)
Câu a
\(S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{2019-2017}{2017x2019}.\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}=1-\frac{1}{2019}=\frac{2018}{2019}\)
Câu b
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\)
\(2A=3A-A=1-\frac{1}{3^7}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^7}\)
a) Ta có: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}=1\text{-}\frac{1}{3}+\frac{1}{3}\text{-}\frac{1}{5}+...+\frac{1}{11}\text{-}\frac{1}{13}=1\text{-}\frac{1}{13}=\frac{12}{13}\)
Thay vào ta có:
\(\frac{12}{13}+x=\frac{24}{13}\Rightarrow x=\frac{24}{13}\text{-}\frac{12}{13}\Rightarrow x=\frac{12}{13}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\right)\)
\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}\right)\)
\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{7}\right)=\frac{5}{2}\left(\frac{7}{7}-\frac{1}{7}\right)=\frac{5}{2}.\frac{6}{7}=\frac{15}{7}\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
Tích mk nha bn !!!! ^_^
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
Chúc bn hok giỏi !!!!!!!!! ^_^
2016/2017 nhé
k cho mình nha
cảm ơn bạn