Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
=> \(2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{16}{34}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
=> \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
=> \(1-\frac{1}{x+2}=\frac{16}{17}\)
=> \(\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
=> \(x+2=17\)
=> \(x=15\)
=>1/1-1/3+1/3-1/5+1/5-1/7+....+1/x-1/(x+2)=16/34
=>1/1-1/(x+2)=16/34
=>1/(x+2)=1-16/34
=>1/(x+2)=9/17
=>(x+2).9=17
=>(x+2)=17/9
=>x=17/9-2
=>x=-1/9(không là số tự nhiên)
vậy không có số tự nhiên x thoả mãn điều kiện bài toán
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2A=1-\frac{1}{2n+1}\)
\(2A=\frac{2n+1-1}{2n+1}\)
\(2A=\frac{2n}{2n+1}\)
\(A=\frac{2n}{2\left(2n+1\right)}\)
\(A=\frac{n}{2n+1}< \frac{n}{2n}=\frac{1}{2}\left(đpcm\right)\)
a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
<=> \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)
<=>\(\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> \(x-2012=0=>x=2012\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)
=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(\frac{2x}{2x+1}=\frac{98}{99}\)
=>2x = 98
=>x = 49
a ) \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
\(< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)=\frac{1}{4}\left(1+\frac{1}{1}-\frac{1}{n}\right)< \frac{1}{2}\)
b )
\(B=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{3^2-1}+\frac{1}{5^2-1}+...+\frac{1}{\left(2n+1\right)^2-1}\)
\(=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n\left(2n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n+2}\right)< \frac{1}{4}\).
đăng vui à :v
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)\)
\(=\frac{1}{2}\left(\frac{2n+1}{2n+1}-\frac{1}{2n+1}\right)\)
\(=\frac{1}{2}\cdot\frac{2n}{2n+1}=\frac{n}{2n+1}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right]\)
\(=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)\)
\(=\frac{n}{2n+1}\)