2.3.4.5 + 4.6.8.10 + 6.9.12.15 / 2.3.4.5 + 6.8.10.12 + 9.12.15.18
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HM
2
LB
9 tháng 9 2018
\(A=\dfrac{2.3.4.5+4.6.8.10+6.9.12.15}{2.3.4.5+6.8.10.12+9.12.15.18}\)
\(A=\dfrac{2.3.4.5+2.2.2.3.2.4.2.5.+3.2.3.3.3.4.3.5}{2.3.4.5+2.2.2.3.2.4.2.5+3.3.3.4.3.5.3.6}\)
\(A=\dfrac{2.3.4.5+2^4.2.3.4.5.+3^4.2.3.4.5}{2.3.4.5+2^4.2.3.4.5+3^4.3.4.5.6}\)
\(A=\dfrac{\left(1+2^4+3^4\right)2.3.4.5}{\left(2+2^5+3^4.6\right)3.4.5}\)
\(A=\dfrac{\left(1+2^4+3^4\right)2}{\left(1+2^4+3^5\right).2}\)
\(A=\dfrac{1+2^4+3^4}{1+2^4+3^5}=\dfrac{1+16+81}{1+16+243}=\dfrac{49}{130}\)
28 tháng 6 2018
\(\dfrac{2.3.4.5+4.6.8.10+6.9.12.15}{2.3.4.5+6.8.10.12+9.12.15.18}\)
\(=\dfrac{4.6.8.10+6.9.12.15}{6.8.10.12+9.12.15.18}\)
\(=\dfrac{1}{3}\)
Kết quả là \(\dfrac{1}{3}\)vì: \(6.8.10.12:4.6.8.10=12:4=3\)
\(9.12.15.18:6.9.12.15=18:6=3\)
Khi cộng vào thì thương không thay đổi và bằng \(\dfrac{1}{3}\)
NL
1
20 tháng 8 2016
Đặt tổng trên là A có:
A = 2.4.6.8 +..+ 100.102.104.106
10A = 2.4.6.8.(10- 0) + 4.6.8.10.(12-2) +....+100.102.104.106.(108-98)
10A= 2.4.6.8.10 + 4.6.8.10.12 -2.4.6.8.10 +....+ 100.102.104.106.108 -98.100.102.104.106
10A= 100.102.104.106.108
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
AH
Akai Haruma
Giáo viên
12 tháng 7 2023
Lời giải:
$A=10(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{92.93.94.95})$
$3A=10(\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{95-92}{92.93.94.95})$
$=10(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{92.93.94}-\frac{1}{93.94.95})$
$=10(\frac{1}{1.2.3}-\frac{1}{93.94.95})$
$A=\frac{10}{3}(\frac{1}{1.2.3}-\frac{1}{93.94.95})$
12 tháng 7 2023
\(A=\dfrac{10}{1.2.3.4}+\dfrac{10}{2.3.4.5}+...+\dfrac{10}{92.93.94.95}\)
\(A=10.\left(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{92.93.94.95}\right)\)
\(3A=10.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{92.93.94.95}\right)\)
\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{92.93.94}-\dfrac{1}{93.94.95}\right)\)
\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{93.94.95}\right)\)
\(3A=10.\left(\dfrac{138415-1}{93.94.95}\right)=\dfrac{1384140}{93.94.95}\)
\(A=\dfrac{461380}{93.94.95}=\dfrac{46138}{93.47.19}=\dfrac{46138}{83049}\)
\(\)
NL
1
BD
5 tháng 8 2023
\(A=\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+\dfrac{1}{3\cdot4\cdot5\cdot6}+....+\dfrac{1}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+\dfrac{3}{3\cdot4\cdot5\cdot6}+...+\dfrac{3}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{9\cdot10\cdot11}-\dfrac{1}{10\cdot11\cdot12}\)\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{10\cdot11\cdot12}\)
\(A=\dfrac{1}{2}-\dfrac{1}{440}\)
\(A=\dfrac{219}{440}\)
\(\frac{2.3.4.5+4.6.8.10+6.9.12.15}{2.3.4.5+6.8.10.12+9.12.15.18}\)
= \(\frac{2.3.4.5+2.3.4.5.16+2.3.4.5.81}{2.3.4.5+2.3.4.5.48+2.3.4.5.243}\)
= \(\frac{2.3.4.5.\left(1+16+81\right)}{2.3.4.5.\left(1+48+243\right)}\)
= \(\frac{98}{292}\)
= 49/146
=2.3.4.5+4.6.8.10+6.9.15/2.3.4.5+6.8.10.3.4+9.12.15.6.3
=2.3.4.5+4.6.8.10+6.9.15/3.(2.3.4.5+9.12.15.6)
=1/3
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