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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....
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow A=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Vậy \(B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
\(\Rightarrow B-A=\frac{1}{1008}\)
=1-1/2+1/2-1/3+1/3-1/4+1*4-1/5+1/5+1/6=1-1/6=5/6
ok xong
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=1-\frac{1}{2}+\)+.... \(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
sách 6,7,8 có 2 bài này nè. mk k bt ghi ps nên mk ko gửi đc sorry nha. Hhh
a)\(A=\frac{10^{2014}+2016}{10^{2015}+2016}=>10A=\frac{10^{2015}+20160}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\left(1\right)\)
\(B=\frac{10^{2015}+2016}{10^{2016}+2016}=>10B=\frac{10^{2016}+20160}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2106}\left(2\right)\)
từ 1 zà 2
=> 10A>10B
=>A>B
A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)
B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
B-A=1/1008