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\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
=> S = P => (S - P)2013 = 0
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1006}\)
\(\Rightarrow S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)\(=P\)
\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Tík cho mik nha!
\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+....+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)}\)
\(=\frac{1}{2014}\)
TRƯỚC TIÊN BẠN NÊN XÉT MẪU SỐ :
*Xét mẫu số :
\(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
\(=\left(1+1+1+...+1\right)+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
(2012 SỐ HẠNG 1) (2012 SỐ HẠNG LÀ PHÂN SỐ)
\(=\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+\left(1+\frac{2012}{4}\right)+...+\left(1+\frac{1}{2013}\right)\)
(2012 SỐ HẠNG (NHÓM))
\(=\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+.....+\frac{2014}{2013}=2014\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}=\frac{1}{2014}\)
Biển Cửa Lò, chùa Thiên mụ, núi Ngũ Hành Sơn, chùa Cầu Hội An, kinh thành Huế, đèo Hải Vân
🐼🐼🐼
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...
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
\(\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}\)
\(=1+\frac{1}{2013}+1+\frac{1}{2012}+1+\frac{1}{2011}+1-\frac{3}{2014}\)
\(=4+\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2014}-\frac{1}{2014}-\frac{1}{2014}\right)\)
Ta có:
\(\frac{1}{2011}>\frac{1}{2014}\Rightarrow\frac{1}{2011}-\frac{1}{2014}>0\)
\(\frac{1}{2012}>\frac{1}{2014}\Rightarrow\frac{1}{2012}-\frac{1}{2014}>0\)
\(\frac{1}{2013}>\frac{1}{2014}\Rightarrow\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Rightarrow\frac{1}{2011}-\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Rightarrow4+\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2014}-\frac{1}{2014}-\frac{1}{2014}\right)>4\)( thêm 2 vế với 4 )
\(\Rightarrow\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}>4\)
Vậy \(\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}>4\)
Tham khảo nhé~
\(D=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
2013/2014