LP
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TU
0
L
1
TU
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
1 tháng 4 2021
\(S=\frac{101}{120}+\frac{1}{2.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{19-18}{18.19}+\frac{20-19}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{20}\right)=\frac{101}{120}+\frac{19}{120}=\frac{120}{120}=1\)
TU
6
6 tháng 5 2016
= 2.(1 / 2.3 + 1 / 3.4 + ..... + 1 / x (x + 1) = 2007/2009
= 2.(1/2 - 1/3 + 1/3 - +.......+ 1/x - 1/x+1) = 2007/2009
= 2.( 1/2 - 1/x+1) = 2007/2009
= 1 - 1/x+1 =2007/2009
= 1/x+1 = 1/2009
=> x + 1 = 2009
=> x = 2008
6 tháng 5 2016
Ta có: 2/2.3 + 2/3.4 + .... + 2/x.(x+1) = 2007/2009
=> 2.[1/2.3+1/3.4+.....+1/x.(x+1)]=2007/2009
=> 2.(1/2-1/3+1/3-1/4 + .... + 1/x - 1/x+1) = 2007/2009
=> 2.(1/2-1/x+1)=2007/2009
=>1/2 - 1/x+1 = 2007/2009 : 2
=> 1/2 - 1/x+1 = 2007/4018
=> 1/x+1 = 2007/4018 +1/2
=> 1/x+1 =
NT
0
NB
6
16 tháng 3 2017
A= 1/4 ( 1/2x6 +1/6x10 +.............+1/46x50)
A= 1/4 ( 1/2 - 1/6 + 1/6 - 1/10 +.......... + 1/46 - 1/50 )
A= 1/4 ( 1/2 - 1/50 )
A= 1/4 x 12/25
A= 3/25
KB
2
7 tháng 7 2021
\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\dfrac{7}{18}:\dfrac{1}{2}\)
\(=\dfrac{7}{9}\)
VA
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 3 2017
Đặt tổng trên là A
\(5A=\frac{5}{4x9}+\frac{5}{9x14}+\frac{5}{14x19}+...+\frac{5}{44x49}\)
\(5A=\frac{9-4}{4x9}+\frac{14-9}{9x14}+\frac{19-14}{14x19}+...+\frac{49-44}{44x49}\)
\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\)
\(5A=\frac{1}{4}-\frac{1}{49}\Rightarrow A=\frac{49-4}{4x5x49}=\frac{45}{4x5x49}=\frac{9}{4x49}\)
\(2023-\dfrac{1}{2\cdot6}-\dfrac{1}{4\cdot9}-...-\dfrac{1}{36\cdot57}-\dfrac{1}{38\cdot60}\)
\(=2023-\dfrac{1}{6}\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{19\cdot20}\right)\)
\(=2023-\dfrac{1}{6}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2023-\dfrac{1}{6}\left(1-\dfrac{1}{20}\right)=2023-\dfrac{1}{6}\cdot\dfrac{19}{20}\)
\(=2023-\dfrac{19}{120}=\dfrac{242741}{120}\)