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\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
\(B=\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
\(=\frac{1}{3022}\left(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+...+\frac{3022}{2014.1008}\right)\)
\(=\frac{1}{3022}\left(\frac{1008}{1008.2014}+\frac{2014}{1008.2014}+...+\frac{2014}{1008.2014}+\frac{1008}{1008.2014}\right)\)
\(=\frac{1}{3022}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+...+\frac{1}{2014}+\frac{1}{1008}\right)\)
\(=\frac{2}{3022}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
=> \(\frac{A}{B}=\frac{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
Vậy....
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
a , Ta có : \(1-\frac{54}{59}=\frac{5}{59}\) \(=\frac{50}{590}\) ; \(1-\frac{541}{591}=\frac{50}{591}\)
Vì \(\frac{50}{590}>\frac{50}{591}\)nên \(\frac{54}{59}< \frac{541}{591}\)
Mình nghĩ là bạn chép nhầm đề vì nếu là vô số số 1 thì không thể tính được. Đề đúng phải là:
Cho \(A=\frac{2016^2+1^2}{2016.1}+\frac{2015^2+2^2}{2015.2}+...+\frac{1009^2+1008^2}{1009.1008}\); \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\)
Tính \(\frac{A}{B}\)
Ta có: \(A=\frac{2016^2+1^2}{2016.1}+\frac{2015^2+2^2}{2015.2}+...+\frac{1009^2+1008^2}{1009.1008}\)
\(=\frac{2016}{1}+\frac{1}{2016}+\frac{2015}{2}+\frac{2}{2015}+...+\frac{1009}{1008}+\frac{1008}{1009}\)
\(=\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}\)
\(=1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)\)
\(=1+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}\)
\(=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=2017\)
Xem kỹ là số
\(B=\frac{1+1+...+1}{2+3+...+2016}\) hay \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\) nhé b
Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)
\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)
\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)
\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)
\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có :
\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)
\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)
\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)
\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)
Mà :
\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)
Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế )
\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 )
\(\Rightarrow\)\(A>3\) ( điều phải chứng minh )
Vậy \(A>3\)
Chúc đệ học tốt ~
c,
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
vì \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.............................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
bt lm mỗi một câu :v
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)