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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2008\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
x+1=2009
x=2009-1=2008
Vậy x bằng 2008
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
Em tham khảo nhé!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}=\frac{2009}{4018}-\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
=> \(1\cdot2009=1\left(x+1\right)\)
=> \(x+1=2009\Rightarrow x=2009-1=2008\)
Vậy x = 2008
Chúc bn hk tốt !
Ta có :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2007}{2009}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2009-1\)
\(\Rightarrow x=2008\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x\left(x+1\right)}\)) \(=\frac{2007}{2009}\)
\(\Rightarrow\)(\(\frac{1}{2}-\frac{1}{x+1}\))\(=\frac{2007}{2009}:2\)
\(\Rightarrow\frac{-1}{x+1}=\frac{2007}{4018}-\frac{1}{2}\)
\(\Rightarrow\frac{-1}{x+1}=\frac{-1}{4018}\Rightarrow x+1=4018\Rightarrow x=4017\)
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{2007}{2009}\)
=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(1-\frac{2}{x+1}=\frac{2007}{2009}\)
=> \(\frac{2}{x+1}=\frac{2}{2009}\) => x + 1 = 2009 => x = 2008
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)