tìm \(\frac{M}{N}\)biết: M=\(\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)và N=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}\)
Những câu hỏi liên quan
1. So sánh M frac{1}{1cdot2}+frac{1}{3cdot4}+...+frac{1}{37cdot38} và N frac{1}{20}+frac{1}{21}+...+frac{1}{38}2. Cho : 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}P frac{1}{1007}+frac{1}{1008}+...+frac{1}{2013} Tìm ( S - P )2013
Đọc tiếp
1. So sánh
M = \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}\) và N = \(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{38}\)
2. Cho :
S = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
P = \(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
Tìm ( S - P )2013
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...-\frac{1}{2012};\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\)Tính \(\left(\frac{A}{B}\right)^{2013}\)
Câu 1: Rút gọn: \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
Câu 2: Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\) và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
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...
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Tự mình trả lời thì đăng câu hỏi làm gì ngáo đá😂😂😂
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Xem thêm câu trả lời
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
Và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
Tính \(\left(S-P\right)^{2013}\)
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
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frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
va P=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
tinh (S-P)2013
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)
\(=P\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0\)
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Cho mình hỏi sao lại trừ 2 lần (1/2 - 1/4 ....) thế ạ
Dạ thôi mình biết rồi ạ
Xem thêm câu trả lời
Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(P=\frac{1}{1007}+\frac{1}{1008}+....+\frac{1}{2012}+\frac{1}{2013}\)
Tính (S-P)2016
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+.....+\frac{1}{2012}+\frac{1}{2013}=P\)
=>S-P=0
=>(S-P)2016=0
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\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
Tính\(\left(S-P\right)^{2013}\)