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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

\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)

\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)

\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)

\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)

\(A=\frac{2013}{2014}\)

\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)

    \(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)

    \(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)

 \(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)         

 \(=\frac{2013}{2014}\)

Cho A=$\frac{n-2}{n+3}$n−2n+‍3 .Tìm giá trị của n để

a) A là phân số

b) A là một số nguyên

mọi người giải hộ tui với!!!

A=\(\frac{n-2}{n+3}\)

\(A=\frac{n-2}{n+3}=\frac{n+3-5}{n+3}=1+\frac{5}{n+3}\)

a) \(n\in R\)  trừ   \(n=-3\)

b) để A là số nguyên thì \(\frac{5}{n+3}\in Z\) 

Suy ra \(n+3\in\left\{1;5;-1;-5\right\}\)

Suy ra \(n\in\left\{-2;2;-4;-8\right\}\)

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

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

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

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

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

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

\(=P\)

\(\Leftrightarrow S-P=0\)

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

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

Dạ thôi mình biết rồi ạ