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\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{7\cdot9}+\dfrac{1}{6\cdot8}\)
\(=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{7\cdot9}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{9}{9}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}+\dfrac{1}{2}\cdot\dfrac{3}{8}\)
\(=\dfrac{1}{2}\left(\dfrac{8}{9}+\dfrac{3}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{64}{72}+\dfrac{27}{72}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{91}{72}\)
\(=\dfrac{91}{144}\)
S=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+...+\dfrac{1}{6.8}\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{2}+...+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{8-1}{8}\right)\)
S=\(\dfrac{1}{2}.\dfrac{7}{8}\)
S=\(\dfrac{7}{16}\)
Giải:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}.\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right).\)
\(=\dfrac{1}{2}\left[\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)+\left(1-\dfrac{1}{2011}\right)\right].\)
\(=\dfrac{1}{2}\left[0+0+0+...+\left(1-\dfrac{1}{2011}\right)\right].\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right).\)
\(=\dfrac{1}{2}.\dfrac{2010}{2011}=\dfrac{2010}{4022}=\dfrac{1005}{2011}.\)
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1\1-1\3+1\3-1\5+1\5-1\7+...+ 1\2009- 1\2011
=1- 1\2011
=2010\2011
dấu \ là 1 trên 
Tính
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + .......+ \(\dfrac{1}{2009.2011}\)
Lời giải:
Ta có: \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2009.2011}\)
\(2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{2011-2009}{2009.2011}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....+\frac{1}{2009}-\frac{1}{2011}\)
\(2A=1-\frac{1}{2011}=\frac{2010}{2011}\Rightarrow A=\frac{1005}{2011}\)
a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
\(B=\dfrac{1}{1.3}\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2003.2005}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\\ =\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}.\dfrac{2004}{2005}\\ =\dfrac{1002}{2005}\)
Help me !!!
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)
\(A=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)
\(A=\dfrac{1}{2}.\dfrac{2010}{2011}\)
\(A=\dfrac{1005}{2011}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2009.2011}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2009.2011}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}.\dfrac{2010}{2011}\)
\(=\dfrac{1005}{2011}\)
Vậy \(A=\dfrac{1005}{2011}\)
a)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}\)
\(=\dfrac{4}{25}\)
b)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)
tương tự
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)......\left(1+\dfrac{1}{2009.2011}\right)\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}........\dfrac{4040100}{2009.2011}\)
\(=\dfrac{2.2.3.3.4.4.....................2010.2010}{1.3.2.4.3.5.................2008.2010.2009.2011}\)
\(=\dfrac{2.2010}{2011}\)
không biết đúng k nhưng cứ lm thôi