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\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
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
Đặt \(A=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)
\(1A=1-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}\right)-\dfrac{1}{22}\)\(1A=1-\dfrac{1}{22}\)
\(1A=\dfrac{22}{22}-\dfrac{1}{22}\)
\(1A=\dfrac{21}{22}\)
\(\dfrac{21}{22}\) không thể rút gọn
\(A=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{6}{16\cdot22}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\\ =1-\dfrac{1}{22}\\ =\dfrac{21}{22}\)
Vậy \(A=\dfrac{21}{22}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{29}\)
=1-1/29
=28/29
sửa đề: \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}\)
giải:
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{22}-\dfrac{1}{25}\\ =1-\dfrac{1}{25}=\dfrac{24}{25}\)
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)
\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{43\cdot46}\\ =\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\\ =1-\dfrac{1}{46}\\ =\dfrac{45}{46}\)
\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\\ =\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{49\cdot51}\\ =\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\cdot\dfrac{16}{51}\\ =\dfrac{8}{51}\)
\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{98\cdot99\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{98\cdot99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}-\dfrac{1}{99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{99\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\\ =\dfrac{1}{2}\cdot\dfrac{4949}{9900}=\dfrac{4949}{19800}\)
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...........+\dfrac{1}{x\left(x+3\right)}=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...........+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+............+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow1-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\dfrac{1}{x+3}=1-\dfrac{18}{19}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{19}\)
\(\Rightarrow x+3=19\)
\(\Rightarrow x=19-3\)
\(\Rightarrow x=16\)
Vậy \(x=16\) laf giá trị cần tìm
A=1/15-1/16+1/16-1/17+...+1/2016-1/2017
A=1/15-1/2017
A=2002/30255
C=1/3[3/5.8+3/8.11+...+3/101.104]
C=1/3[1/5-1/8+1/8-1/11+...+1/101-1/104]
C=1/3[1/5-1/104]
C=1/3.99/520
C=33/520
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
\(A=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot11}+\dfrac{1}{11\cdot14}+\dfrac{1}{14\cdot17}\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{17}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{16}{17}=\dfrac{16}{51}\)