Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(1-\frac{1}{3}\right)\)\(x\left(1-\frac{1}{6}\right)\)\(x\)\(\left(1-\frac{1}{10}\right)\)\(x\)\(\left(1-\frac{1}{15}\right)\)\(x\)\(\left(1-\frac{1}{21}\right)\)\(x\)\(\left(1-\frac{1}{28}\right)\)\(=\)\(\left(\frac{3}{3}-\frac{1}{3}\right)\)\(x\)\(\left(\frac{6}{6}-\frac{1}{6}\right)\)\(x\)\(\left(\frac{10}{10}-\frac{1}{10}\right)\)\(x\)\(\left(\frac{15}{15}-\frac{1}{15}\right)\)\(x\)\(\left(\frac{21}{21}-\frac{1}{21}\right)\)\(x\)\(\left(\frac{28}{28}-\frac{1}{28}\right)\)\(=\)\(\frac{2}{3}x\frac{5}{6}x\frac{9}{10}x\frac{14}{15}x\frac{20}{21}x\frac{27}{28}\)\(=\)\(\frac{2x5x9x14x20x27}{3x6x10x15x21x28}\)\(=\)\(\frac{2x5\left(3x3\right)x\left(2x7\right)x\left(5x4\right)x\left(3x3x3\right)}{3x\left(3x2\right)x\left(5x2\right)x\left(5x3\right)x\left(7x3\right)x\left(4x7\right)}\)\(=\)\(\frac{3}{7}\)
=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2013}{2014}x\frac{2014}{2015}\)
=\(\frac{1x2x3x...x2013x2014}{2x3x4x...x2014x2015}\)
=\(\frac{1}{2015}\)
( Dau x la dau nhan)
\(P=\frac{1}{5x8}+\frac{1}{8x11}+.....+\frac{1}{602x605}\)
\(\Rightarrow3P=\frac{3}{5x8}+\frac{3}{8x11}+......+\frac{3}{602x605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.....+\frac{1}{602}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{24}{121}\)
\(\Rightarrow P=\frac{24}{121}:3\)
\(\Rightarrow P=\frac{8}{121}\)
a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'
= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009
= 2010/2005
= 402/401
\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)
\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)
\(=\frac{2010}{2005}\)
\(=\frac{402}{401}\)