chung to A=\(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2013!}<2\)
(n!=1.2.3....n duoc goi la giai thua)
Cho A = \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2012^2}+\frac{1}{2013^2}\)
Hay Chung to rang A < 1
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Ta có : A= 1/2^2 +1/3^2 +....+1/2012^2 +1/2013^2
=> A= 1/2.2 +1/3.3 +....+1/2012.2012 +1/2013.2013
Do :1/2.2< 1/1.2
1/3.3 <1/2.3
.................
1/2012.2012 <1/2011.2012
1/2013.2013< 1/2012.2013
=>1/2.2 +1/3.3 +...+1/2012.2012+1/2013.2013< 1/1.2 +1/2.3+...+1/2011.2012+1/2012.2013
=>A<1/1 -1/2 +1/2 -1/3+...+1/2011-1/2012+1/2012-1/2013
=>A<1/1-1/2013
=>A<2013/2013 -1/2013
=> A< 2012/2013
Vì 2012<2013=>2012/2013<1
mà A<2012/2013=>A<1
Vậy A<1
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2012^2}< \frac{1}{2011.2012};\frac{1}{2013^2}< \frac{1}{2012.2013}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}+\frac{1}{2012.2013}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow A< 1-\frac{1}{2013}\)
Mặt khác : \(1-\frac{1}{2013}< 1\)
\(\Rightarrow A< 1-\frac{1}{2013}< 1\)
\(\Rightarrow A< 1\)
\(\)
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Tính giá trị biểu thức :
\(A=\frac{\frac{1}{2013}+\frac{2}{2012}+\frac{3}{2011}+...+\frac{2011}{3}+\frac{2012}{2}+\frac{2013}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
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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 : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
Tính giá trị biểu thức :
\(A=\frac{\frac{1}{2013}+\frac{2}{2012}+\frac{3}{2011}+...+\frac{2011}{3}+\frac{2012}{2}+\frac{2013}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
Ta có: Tử là:
B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\) (2013 số hạng 1)
=\(\left(\frac{1}{2013}+1\right)+\left(\frac{2}{2012}+1\right)+...+\left(\frac{2012}{2}+1\right)+\left(1\right)\)
=\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)
=\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014
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cho \(a+b+c=2013\)va\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2013}\).
Chung minh rang trong 3 so co it nhat 1 so bang 2013
A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+.....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}+\frac{1}{2015}}\)=
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
toán lớp 8 dễ quá vậy
A=2015
hình như thế
chung minh
\(\frac{1}{2.2}+\frac{1}{2.3}+...+\frac{1}{2013}+\frac{1}{2013}\)
giá trị của biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)