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 \(n\in\left\{-2;2;-4;-8\right\}\)

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

óc chó      c hó

B=2013.(1+

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

B=2013(\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)

B=2013.2(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)

Maỳ có bị óc chó không mà bảo câu trả lời của đại ca tao là sai

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

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

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

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

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

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

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

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

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

 \(=\frac{2013}{2014}\)