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\(=\frac{3999999-1}{1998+3998000}=\frac{3999998}{3999998}=1\)
\(\frac{1999x2001-1}{1998+1999x2000}\)
\(=\frac{1999x2000+1999-1}{1998+1999x2000}\)
\(=\frac{1999x2000+1998}{1998+1999x2000}\)
\(=1.\)
Đúng 100%
\(\frac{1999\times2001-1}{1999+1999\times2000}\times\frac{7}{5}\)
\(=\frac{2001-1}{1999\times2000}\times\frac{7}{5}\)
\(=\frac{2000}{1999\times2000}\times\frac{7}{5}\)
\(=1999\times\frac{7}{5}\)
........... còn lại tự xử lý nha
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(\dfrac{1999\times2001-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)
= \(\dfrac{1999\times\left(2000+1\right)-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)
= \(\dfrac{1999\times2000+1999-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)
= \(\dfrac{1999\times2000+1998}{1999\times2000+1998}\) \(\times\) \(\dfrac{7}{5}\)
= 1 \(\times\) \(\dfrac{7}{5}\)
= \(\dfrac{7}{5}\)