A= 10/1.2.3.4+10/2.3.4.5+...+10/92.93.94.95
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12 tháng 7 2023
\(A=\dfrac{10}{1.2.3.4}+\dfrac{10}{2.3.4.5}+...+\dfrac{10}{92.93.94.95}\)
\(A=10.\left(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{92.93.94.95}\right)\)
\(3A=10.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{92.93.94.95}\right)\)
\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{92.93.94}-\dfrac{1}{93.94.95}\right)\)
\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{93.94.95}\right)\)
\(3A=10.\left(\dfrac{138415-1}{93.94.95}\right)=\dfrac{1384140}{93.94.95}\)
\(A=\dfrac{461380}{93.94.95}=\dfrac{46138}{93.47.19}=\dfrac{46138}{83049}\)
\(\)
4 tháng 6 2019
1.
A= 5/28 + 5/70 +.....+10/700 = 5/(4.7)+5/(7.10)+....5/(25.28)
3A= 5( 1/4 - 1/7 +1/7-1/10+......+1/25-1/28)
3A= 5 (1/4-1/28)
3A=15/14
A= 5/14
T
4 tháng 6 2019
#)Giải :
1. \(A=\frac{10}{54}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(A=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(\Rightarrow\frac{3A}{5}=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\Rightarrow\frac{3A}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\)
\(\Rightarrow\frac{3A}{5}=\frac{1}{4}-\frac{1}{28}=\frac{3}{14}\)
\(\Rightarrow A=\frac{3}{14}\times\frac{5}{3}\)
\(\Rightarrow A=\frac{5}{14}\)
KT
1
BT
12 tháng 3 2018
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)
=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)
=> \(A=\frac{451}{8120}\)
S
13 tháng 10 2016
Đặt S=1.2.3.4+2.3.4.5+...+97.98.99.100
5S=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5
5S=1.2.3.4.(5 - 0)+2.3.4.5.(6 - 1)+...+97.98.99.100.(101 - 96)
5S=1.2.3.4.5-0.1.2.3.4+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99
5S=97.98.99.100.101
S=97.98.99.20.101
=>S=1901009880
13 tháng 10 2016
Đặt A = 1.2.3.4 + 2.3.4.5 + ... + 97.98.99.100
5A = 1.2.3.4.5 + 2.3.4.5.5 + ... + 97.98.99.100.5
5A = 1.2.3.4.5 + 2.3.4.( 6 - 1 ) + ... + 97.98.99.100.( 101 - 96 )
5A = 1.2.3.4.5 + 2.3.4.5.6 - 1.2.3.4.5 + ... + 97.98.99.100.101 - 96.97.98.99.100
5A = 97.98.99.100.101
A = 97.98.99.100.101 : 5
A = 97.98.20.101
A = 19202120
QD
5 tháng 6 2016
Đặt A=1.2.3.4+2.3.4.5+...+97.98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
A=98.99.100.101/4
CH
5 tháng 6 2016
5A=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100
5A=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5A=97.98.99.100.101=9505049400
A=1901009880
NL
1
BD
5 tháng 8 2023
\(A=\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+\dfrac{1}{3\cdot4\cdot5\cdot6}+....+\dfrac{1}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+\dfrac{3}{3\cdot4\cdot5\cdot6}+...+\dfrac{3}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{9\cdot10\cdot11}-\dfrac{1}{10\cdot11\cdot12}\)\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{10\cdot11\cdot12}\)
\(A=\dfrac{1}{2}-\dfrac{1}{440}\)
\(A=\dfrac{219}{440}\)
10 tháng 6 2016
\(\text{Ta có S=1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100
}\)
\(\text{\Rightarrow5S=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101- 96).97.98.99.100
}\)
\(\text{\Rightarrow5S=1.2.3.4.5-0+2.3.4.5.6- 1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
}\)
5S=97.98.99.100.101
5S =9505049400
10 tháng 6 2016
Đặt S= 1.2.3.4+2.3.4.5.6+3.4.5.6+...+97.98.99.100
5S=(5-0).1.2.3.4+(6-1)+2.3.4.5+...+(101-96).97.98.99.100
5S=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5S=97.98.99.100.101=9505049400
S=1901009880
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HH
0
Lời giải:
$A=10(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{92.93.94.95})$
$3A=10(\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{95-92}{92.93.94.95})$
$=10(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{92.93.94}-\frac{1}{93.94.95})$
$=10(\frac{1}{1.2.3}-\frac{1}{93.94.95})$
$A=\frac{10}{3}(\frac{1}{1.2.3}-\frac{1}{93.94.95})$