tim tich: A= (1+1/1 x 3)x(1+1/2x4)x(1+1/3x5)x............x(1+1/2011x2013)
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NL
1
12 tháng 4 2017
Ta có :
\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).................\left(1+\dfrac{1}{2011.2013}\right)\)
\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\left(\dfrac{8}{8}+\dfrac{1}{8}\right)................\left(\dfrac{9999}{9999}+\dfrac{1}{9999}\right)\)
\(A=\dfrac{4}{3}.\dfrac{9}{8}...............\dfrac{10000}{9999}\)
\(A=\dfrac{4.9.................10000}{3.8.............9999}\)
\(A=\dfrac{2.2.3.3................100.100}{1.3.2.4...............99.101}\)
\(A=\dfrac{2.100}{101}=\dfrac{200}{101}\)
~ Chúc bn học tốt ~
LN
0
9 tháng 8 2016
A= (1+1/1 x 3)x(1+1/2x4)x(1+1/3x5)x............x(1+1/2011x2013)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)....\left(\frac{4048143}{4048143}+\frac{1}{4048143}\right)\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot...\cdot\frac{4048144}{4048143}\)
\(=\frac{4\cdot9\cdot....\cdot4048144}{3\cdot8\cdot....\cdot4048143}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot....\cdot2012\cdot2012}{1\cdot3\cdot2\cdot4\cdot....\cdot2011\cdot2013}\)
\(=\frac{2\cdot2012}{2013}=\frac{4024}{2013}\)
NT
2
AH
Akai Haruma
Giáo viên
15 tháng 8 2023
Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Khi đó:
$1+\frac{1}{1.3}=\frac{2^2}{1.3}$
$1+\frac{1}{2.4}=\frac{3^2}{2.4}$
.........
$1+\frac{1}{99.101}=\frac{100^2}{99.101}$
Khi đó:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
ND
3
DK
0
TN
0
NK
4
MT
12 tháng 7 2015
\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)
\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)
\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)
\(=\frac{2\times100}{101}=\frac{200}{101}\)
\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2011.2013}\right)\)
\(A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{4048144}{2011.2013}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2012.2012}{2011.2013}\)
\(A=\frac{2.3.4...2012}{1.2.3...2011}.\frac{2.3.4...2012}{3.4.5...2013}\)
\(A=2012.\frac{2}{2013}=\frac{4024}{2013}\)