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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.
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
P= \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.17}\)+...+\(\dfrac{3}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{5}{3}\).(\(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+...+\(\dfrac{3}{96.101}\))
\(\dfrac{5}{3}\).P= \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+...+\(\dfrac{5}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{96}\)-\(\dfrac{1}{101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{101}\)= \(\dfrac{101}{101}\)-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)
P= \(\dfrac{100}{101}\):\(\dfrac{5}{3}\)= \(\dfrac{100}{101}\).\(\dfrac{3}{5}\)=\(\dfrac{100.3}{101.5}\)=\(\dfrac{20.3}{101.1}\)=\(\dfrac{60}{101}\)
Vậy P= \(\dfrac{60}{101}\)
a, \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(A=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(A=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
b, \(B=\dfrac{6}{15.18}+\dfrac{6}{18.21}+...+\dfrac{6}{87.90}\)
\(B=2\left(\dfrac{3}{15.18}+\dfrac{3}{18.21}+...+\dfrac{13}{87.90}\right)\)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)=2.\dfrac{1}{18}=\dfrac{1}{9}\)
c, \(C=\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
\(C=3\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)=3.\dfrac{3}{35}=\dfrac{9}{35}\)
Chúc bạn học tốt!!!
\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)
\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)
\(S=\frac{60}{101}\)
\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)
\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\frac{100}{101}\)
\(=\frac{60}{101}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{6}{31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
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
\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\ S=\dfrac{25}{1.6}+\dfrac{25}{6.11}+\dfrac{25}{11.16}+...+\dfrac{25}{96.101}\\ S=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ S=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ S=5.\left(1-\dfrac{1}{101}\right)\\ S=5.\dfrac{100}{101}\\ S=\dfrac{500}{101}\)
S=500/101
Huỳnh Huyền Linh làm đúng rùi!