Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/1.2+1/2.3+.....+1/x.(x+1)=2008/2009
=>1/1-1/2+1/2-1/3+.....+1/x-1/x+1=2008/2009
=>1/1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/x+1/x)-1/x+1=2008/2009
=>1/1+0+0+.....+0-1/x+1=2008/2009
=>1-1/x+1=2008/2009
=>1/x+1=1-2008/2009=1/2009
=>x+1=2009
=>x=2008
vậy x=2008
\(\Leftrightarrow\frac{1}{2}+\left(\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}+2.\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3}{10}\)
\(\Leftrightarrow2.\left(\frac{1}{7}-\frac{1}{x+1}\right)=\frac{3}{10}-\frac{1}{2}=-\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{7}-\frac{1}{x+1}=-\frac{1}{5}:2=-\frac{1}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{7}-\left(-\frac{1}{10}\right)=\frac{17}{70}\)
\(\Rightarrow17x+17=70\)
=> không tồn tại n vì n là số tự nhiên
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
a) 2\(\frac{x}{7}\) = \(\frac{75}{35}\)
\(\frac{2.7+x}{7}\) = \(\frac{75:5}{35:5}\) = \(\frac{15}{7}\)
=> 2.7+x = 15
14+x = 15
x = 15-14 = 1
Vậy x=1
b)4\(\frac{3}{x}\) = \(\frac{47}{x}\)
\(\frac{4.x+3}{x}\) = \(\frac{47}{x}\)
=> 4.x + 3 = 47
4x= 47-3=44
vậy x= 44:4=11
c)x\(\frac{x}{15}\) = \(\frac{112}{5}\)
x\(\frac{x}{15}\) =\(\frac{112.3}{5.3}\) = \(\frac{336}{15}\)
\(\frac{x.15+x.1}{15}\) = \(\frac{336}{15}\)
=>(15+1) x =336
16x = 336
x = 336 : 16
vậy x = 21
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
1/3.4+1/4.5+1/5.6+.....+1/x(x+1)=3/10
1/3-1/4+1/4-1/5+1/5-........-1/x+1/x-1/x+1=3/10
=>1/3-1/x+1=3/10
1/x+1=3/10-1/3=1/30
=>x+1=30
x=30-1
x=29
Ta có :
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
=>\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
=>\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
=>\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
=>\(\frac{1}{x+1}=\frac{1}{30}\)
=>\(x+1=30\)
=>\(x=30-1\)
=>\(x=29\)
Vậy \(x=29\)
\(\dfrac{x}{2008}+\dfrac{x}{2009}-\dfrac{x}{2007}=1+\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{2}{2007}\)
\(\Rightarrow x = \dfrac{2007.2008.2009+2009.2007-2008.2007-2.2008.2009}{2009.2007+2008.2007-2008.2009}\)
sai rồi