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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2009}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2008}{2009}\)
=\(\frac{1}{2009}\)
(1-1/2)(1-1/3)(1-1/4)...(1-1/2009)
=1/2*2/3*3/4*...*2008/2009
=\(\frac{1\cdot2\cdot3\cdot...\cdot2008}{2\cdot3\cdot4\cdot...\cdot2009}\)
=1/2009
=> 1/2.2/3.3/4...x/(x+1)=1/2009
=> 1/(x+1)=1/2009
=> x+1=2009
=> x=2008
Mik nhanh nhất mik nha
=> 1/2 . 2/3 . 3/4 .... x/x+1 = 1/2009
<=> (1.2.3...x) / [2.3...x.(x+1)] = 1/2009
<=> 1/ (x+1) = 1/2009
=> x+1=2009 => x=2008
trong tích trên có 1 thừa số như thế này:
\(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{125}\right)\)
=0
=> tích trên bằng 0
\(\frac{2011.4023+2012}{2012.4023-2011}=\frac{2011.4023+2011+1}{2012.4023-2012-1}=\frac{2011.4023+2011.1+1}{2012.4023-2012.1-1}\)
\(=>\frac{2012.4023+2012.1+1}{2012.4023-2012.1-1}=\frac{2012.\left(4023+1\right)+1}{2012.\left(4023-1\right)-1}\)
\(=\frac{4023+1+1}{4023-1-1}=\frac{4023+2}{4023-2}=\frac{4025}{4021}\)
Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1
<br class="Apple-interchange-newline"><div id="inner-editor"></div>=>2012.4023+2012.1+12012.4023−2012.1−1 =2012.(4023+1)+12012.(4023−1)−1
=4023+1+14023−1−1 =4023+24023−2 =40254021
Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)
\(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{2006}{2007}.\frac{2007}{2008}\)
\(=\frac{9.10.11.....2006.2007}{10.11.12.....2007.2008}\)
\(=\frac{9}{2008}\)
\(Ta\) \(có:\)
\(A=\frac{9}{2008}\)
\(B=\frac{1}{2000}\)
\(\frac{9}{2008}=\frac{9.250}{2008.250}=\frac{2250}{502000}\)
\(\frac{1}{2000}=\frac{1.251}{2000.251}=\frac{251}{502000}\)
Vì \(\frac{2250}{502000}>\frac{251}{502000}\Rightarrow A>B\)
\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)
\(A=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}....\frac{2006}{2007}.\frac{2007}{2008}\)
\(A=\frac{9.10.11....2006.2007}{10.11.12...2007.2008}\)
\(A=\frac{9}{2008}\)
Vì \(\frac{9}{2008}