Chứng minh rằng : \(A=1.2.3...2010\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)⋮2011\)
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18 tháng 4 2018
Suy ra : A = ( 1 - 1 / 2010 ) . ( 1 - 2 / 2010 ) .... 0 . ( 1 - 2011 / 2010 ) = 0
Suy ra A = 0
18 tháng 4 2018
A = 1. ( 1/2010 + 2/2010 ) - ( 3/2010 + 4/2010 ) - ... - ( 2010/2010 + 2011/2010 )
= 1/2010 - 2011/2010
= -2010/2010
29 tháng 2 2016
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
27 tháng 4 2016
\(A=\frac{ }{ }sdadsad\text{đ}\text{s}gh\text{d}fg\text{d}\)sf
2 tháng 8 2019
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
\(=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...0\left(1-\frac{2011}{2010}\right)\)
\(=0\)
16 tháng 8 2018
ta có \(\left(1-\frac{1}{2010}.\right).\left(1-\frac{2}{2010}\right)....\left(1-\frac{2010}{2010}\right).\left(1-\frac{2011}{2010}\right)\)\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).......0.\left(1-\frac{2011}{2010}\right)=0\)
T
20 tháng 6 2019
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
20 tháng 6 2019
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
6 tháng 5 2017
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
6 tháng 5 2017
\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
Ta có: \(A=1.2.3...2010\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)\)
\(=\)1.2.3...2010\([\left(1+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+...+\left(\frac{1}{1005}+\frac{1}{1006}\right)]\)
\(=\)\(1.2.3...2010\left(\frac{2011}{2010}+\frac{2011}{2009.2}+...+\frac{2011}{1005.1006}\right)\)
\(=2011\left(\frac{2010!}{2010}+\frac{2010!}{2009.2}+...+\frac{2010!}{1005.1006}\right)\)
Suy ra: A ⋮ 2011
Vậy A ⋮ 2011