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

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

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

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

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

à há mình ko biết

 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

Ko hiểu

S = 1/3+1/5+1/7+...+1/2013-(1/2+1/4+1/6+...+1/2012)

S = 1/2+1/3+1/4+...+1/2012+1/2013 - 2(1/2+1/4+1/6+...+1/2012)

S = 1/2+1/3+1/4+...+1/2012+1/2013 - (1+1/2+1/3+...+1/1006)

S = 1/1007+1/1008+...+1/2013-1

=> S - P = 1/1007+1/1008+...+1/2013-1-(1/1007+1/1008+...+1/2013)

<=> S - P= -1 <=> (S-P)²⁰¹³ = -1

ket qua la (-1)

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

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

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

=>(S-P)²⁰¹³=0²⁰¹³=0

cảm ơn pạn nhak Tuấn Minh. Kết bạn nhak.

link này nè bn!

https://olm.vn/hoi-dap/detail/103540952175.html

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

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

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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...

Tự mình trả lời thì đăng câu hỏi làm gì ngáo đá😂😂😂