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1)\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)
\(\Rightarrow4S=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{2003.2005.2007}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\)
\(=\frac{1}{3}-\frac{1}{4024035}=\frac{1341345}{4024035}=\frac{1}{3}\)
\(\Rightarrow S=\frac{1}{3}:4\approx0,08\)
2)\(S=\frac{1}{3}:4=\frac{1}{12}\)
\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)
\(S=\frac{2}{2}.\frac{1}{1.3.5}+\frac{2}{2}.\frac{1}{3.5.7}+\frac{2}{2}.\frac{1}{5.7.9}+...+\frac{2}{2}.\frac{1}{2003.2005.2007}\)
\(S=\frac{1}{2}.\frac{2}{1.3.5}+\frac{1}{2}.\frac{2}{3.5.7}+\frac{1}{2}.\frac{2}{5.7.9}+...+\frac{1}{2}.\frac{2}{2003.2005.2007}\)
\(S=\frac{1}{2}.\left(\frac{2}{1.3.5}+\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{2003.2005.2007}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{2005.2007}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4024035}\right)=\frac{1}{2}.\frac{1341345}{4024035}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)
Vậy \(S=\frac{1}{6}\)
B=\(\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+........+\frac{1}{2009+2011+2013}\)
\(\Leftrightarrow4B=\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+.....+\frac{4}{2009+2011+2013}\)
4B=\(\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+\frac{1}{5\cdot7}-\frac{1}{7\cdot9}+......-\frac{1}{2009\cdot2011}+\frac{1}{2009\cdot2011}-\frac{1}{2011\cdot2013}\)
4B=\(\frac{1}{3\cdot5}-\frac{1}{2011\cdot2013}\)=>B=\(\left(\frac{1}{3\cdot5}-\frac{1}{2011\cdot2013}\right)\div4\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
Học tốt nha^^
cau a),b),c) ban dat mau chung roi khu mau ma lam la duoc ma
Tuy học lớp 6 ................. cơ mừ thấy mí bài nỳ dễ quá >.<
A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)
A=1-\(\frac{1}{2018}\)
A=\(\frac{2017}{2018}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)
Đến đây bình thường ta nhóm 2 số vào với nhau nhưng ở đây có lẻ số hạng nên không nhóm được => đề sai
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=9\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=7\)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)
\(=\frac{1}{1.3}-\frac{1}{1999.2001}\)
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