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\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{11}\)
\(S.2=\frac{10}{11}\)
\(S=\frac{10}{11}:2\)
\(S=\frac{5}{11}\)
Tìm x:
\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{19x21}\right).x=\frac{9}{7}\)
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)
\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)
\(\frac{1}{7}.x=\frac{9}{7}\)
\(x=\frac{9}{7}\div\frac{1}{7}\)
\(x=9\)
Vậy ...
\(A=\frac{1}{3}-\frac{1}{17}=\frac{14}{51}\)
cách làm thì tự biết
trên mạng đầy
kết quả đúng phải là 7/51 chứ bn
mk cần cách trình bày thôi
câu trả lời của bn hơi lạnh nhạt tí ^.^
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
\(A=\frac{1}{3}-\frac{1}{19}\)
\(A=\frac{16}{57}\)
Dấu "." là dấu nhân nhá ^^
\(\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{2009\cdot2011}+\frac{1}{x}=\frac{1}{5}\cdot0,5\)
\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{2011-2009}{2009\cdot2011}+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2009}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left(\frac{1}{2}\cdot\frac{2006}{10055}\right)+\frac{1}{x}=\frac{1}{10}\)
\(=\frac{1003}{10055}+\frac{1}{x}=\frac{1}{10}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{10}-\frac{1003}{10055}\)
\(\frac{1}{x}=\frac{1}{4022}\)
\(\Rightarrow x=1\div\frac{1}{4022}=4022\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)