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S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
S-P= 0
(S-P)^2013 = 0

DD
16 tháng 1 2021

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)

\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)

\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\).

22 tháng 3 2021

à há mình ko biết

16 tháng 4 2017

 S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
 S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
 S-P= 0
Suy ra (S-P)^2013 = 0

15 tháng 1 2018

Ko hiểu

19 tháng 7 2018

\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2012}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)

\(=\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}=P\)

Vậy \(S=P\)

10 tháng 4 2017

\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1006}\)

\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}=P\)

=>(S-P)2013=02013=0

10 tháng 4 2017

cảm ơn pạn nhak Tuấn Minh. Kết bạn nhak.

7 tháng 4 2018

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)

\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)

\(=P\)

\(\Leftrightarrow S-P=0\)

\(\Leftrightarrow\left(S-P\right)^{2013}=0\)

20 tháng 3 2020

Cho mình hỏi sao lại trừ 2 lần (1/2 - 1/4 ....) thế ạ

2 tháng 4 2019

Biển Cửa Lò, chùa Thiên mụ, núi Ngũ Hành Sơn, chùa Cầu Hội An, kinh thành Huế, đèo Hải Vân

🐼🐼🐼

2 tháng 4 2019

Ta có:

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)

\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)

Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)

Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)

Vậy...

25 tháng 3 2017

Ta có: \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)

\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)

\(\Rightarrow P-S=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\right)=0\)

\(\Rightarrow\left(P-S\right)^{2013}=0^{2013}=0\)

Vậy \(\left(P-S\right)^{2013}=0\)

30 tháng 4 2018

Hay quá