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=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 = \(\dfrac{502}{3}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)
\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)
\(=\dfrac{3}{4}-\dfrac{5}{12}\)
\(=\dfrac{4}{12}=\dfrac{1}{3}\)
Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.
Ko biết sau này câu trả lời có xuất hiện hay ko...
2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{97}{300}\)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$