TìM X : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....................+\frac{2}{x.\left(x-1\right)}=1\frac{2003}{2005}\)
Tìm số tự nhiên x biết \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)
\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)
=>x+1=2005
=>x=2004
1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015
Đ/A là 2004
chúc đồng chí Chế Minh Hải học tốt
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)
\(\Leftrightarrow4006x+4006=4010x-4010\)
\(\Leftrightarrow-4x=-8016\)
\(\Leftrightarrow x=2004\)
Vậy x = 2004
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)
\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)
\(\left(x+1\right).7=4020\)
\(\Rightarrow x=....\)
Tìm x biết
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
PS : Không cần làm đẹp, càng nhanh càng tốt, quan trọng là đúng!
Ta có :
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)
\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)
\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)
Vậy \(x=\frac{2003}{2007}\)
Chúc bạn học tốt ~
Tìm x
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+.....+\(\frac{2}{x\left[x+1\right]}\)=\(\frac{2003}{2005}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
Tìm x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2007}\)
\(\Rightarrow x+1=2007\)
\(\Rightarrow x=2006\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{2005}{2007}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)
\(\Rightarrow x+1=2007\)
\(x=2007-1\)
\(x=2006\)
1 tìm x thuộc N biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...................+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
Tìm x biết : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..,+\frac{2}{x\left(x+1\right)}=\frac{2004}{2005}\)
Lời giải:
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{2004}{2005}$
$2(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)})=\frac{2004}{2005}$
$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}= \frac{1002}{2005}$
$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$
$\Rightarrow x+1=4010$
$\Rightarrow x=4009$
Tìm x biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2004}{2005}\)
Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2004}{2005}$
$2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}\right]=\frac{2004}{2005}$
$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$
$\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{(x+1)-x}{x(x+1)}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$
$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$
$x+1=4010$
$x=4010-1=4009$