A = (2013-1) (2013-2) (2013-3)...(2013-n)
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biết rằng tích A=(2013-1).(2013-2).(2013-3) ... (2013-n) có đúng 2013 thừa số. khi đó A = ?
Tích trên có 2013 thừa số suy ra n bằng 2013 => 2013 - 2013 = 0 =>A = 0
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Biết rằng tích A=(2013-1).(2013-2).(2013-3).....(2013-n) có đúng 2013 thừa số.Khi đó A bằng....
Có đúng 2013 thừa số
< = > Thừa số cuối cùng là 2013 - 2013 = 0
Tích là:
(2013-1).(2013-2).(2013-3).....0 = 0
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Biết rằng tích A= (2013-1).(2013-2).(2013-3)...(2013-n) có đúng 2013 thừa số . Khi đó A bằng:
A = (2013/2 + 2013/3+2013/4 + ....+2013/2014) : (2013/1+2012/2 +2011/3+...+1/2013)
\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(A=\frac{2013}{2014}\)
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\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
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Biết rằng tích A=(2013-1).(2013-2).(2013-3).........(2013-n)
neu cach giai gium minh nhe
Tính \(A=2013+\frac{2013}{1+2}+\frac{2013}{1+2+3}+\frac{2013}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
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31 tháng 12 2017 lúc 22:12
Tính tổng :
\(S=\frac{1}{2013-1}+\frac{2}{2013+1}+\frac{2^2}{2013^2+1}+\frac{2^3}{2013^{2^2}+1}+.....+\frac{2^{n+1}}{2013^{2^n}+1}\)
2013+(2013/1+2)+(2013/1+2+3)+(2013/1+2+3+4)+...+(2013/1+2+3+...+2012)
\(ChoA=\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2013}\) và B=\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)Tính\(\frac{A}{B}\)