Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\) nên
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)
\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)
Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)
b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)
\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)
\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)
\(\Rightarrow n+1=4\Rightarrow n=3\)
\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)
\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=5\)
c ) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)
\(\Leftrightarrow p=4\)
\(a.\)
\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
Vậy : \(m=4\)
\(b.\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)
\(\Rightarrow n=5\)
Vậy : \(n=5\)
\(c.\)
\(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
Vậy : \(p=4\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
\(A=\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{1973.2003}\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}-\frac{1}{32}-\frac{1}{33}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{31.2003}\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Vậy \(\frac{A}{B}=\frac{1972}{30}\)