(1-1/9) x(1-1/100) x...x(1-1/2006) = ?
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(1-1/99).(1-1/100)....(1-1/2006) =(99/99-1/99).(100/100-1/100)....(2006/2006-1/2006) = (98/99).(99/100)...(2005/2006) =98/2006=..
1-1/99).(1-1/100)....(1-1/2006)
=(99/99-1/99).(100/100-1/100)....(2006/2006-1/2006) = (98/99).(99/100)...(2005/2006) =98/2006=..
\(\left(1-\frac{1}{99}\right)\times\left(1-\frac{1}{100}\right)\times...\times\left(1-\frac{1}{2006}\right)\)
\(=\left(\frac{99}{99}-\frac{1}{99}\right)\times\left(\frac{100}{100}-\frac{1}{100}\right)\times\left(\frac{2006}{2006}-\frac{1}{2006}\right)\)
\(=\frac{98}{99}\times\frac{99}{100}\times...\times\frac{2005}{2006}\)
\(=\frac{98\times99\times...\times2005}{99\times100\times...\times2006}\)
\(=\frac{98}{2006}=\frac{49}{1003}\)
= 49\1003 đung ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
Ta có : \(\left(1\frac{1}{99}\right)\times\left(1\frac{1}{100}\right)\times...\times\left(1\frac{1}{2006}\right)\)
\(=\) \(\frac{100}{99}\times\frac{101}{100}\times...\times\frac{2007}{2006}\)
\(=\) \(\frac{100\times101\times...\times2007}{99\times100\times...\times2006}\)
\(=\) \(\frac{2007}{99}\)
\(=\) \(\frac{223}{11}\)
Đánh máy mệt v~~ Haizzz ...
P/s : Bn cứ lm` theo mk là OK ko cần bàn cãi =))
> Chúc họk tốt <
Bài 1
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2
\(\dfrac{111}{333}=\dfrac{111:111}{333:111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222:2222}{4444:2222}=\dfrac{1}{2}\)
Do \(3>2\Rightarrow\dfrac{1}{3}< \dfrac{1}{2}\)
Vậy \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
Bài 1.
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98\times99\times...\times2005}{99\times100\times...2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2.
Có: \(\dfrac{111}{333}=\dfrac{111}{3\times111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222}{2\times2222}=\dfrac{1}{2}\)
Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) nên \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
a)=(3/8+10/16)+(7/12+10/24)
=1+1=2
c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)
=3+2+2=7
a: \(=\dfrac{\left(\dfrac{1}{2}:\dfrac{1}{2}-\dfrac{1}{4}:\dfrac{1}{4}+\dfrac{1}{8}:\dfrac{1}{8}-\dfrac{1}{10}:\dfrac{1}{10}\right)}{1+2+3+...+2008}\)
=0
c: =8,1*5/3*1875+13,5*625
=13,5(1875+625)
=13,5*2500
=33750