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6 tháng 5 2018
Tớ không chép lại đề nữa nhé:
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)=\(\frac{1}{2}.\left(\frac{3-1}{1-3}+\frac{7-5}{5-7}+...+\frac{2011-2009}{2009-2011}\right)\)
= \(\frac{1}{2}.\left(\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{2011}{2009.2011}-\frac{2009}{2009.2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\frac{2010}{2011}\)
=\(\frac{1005}{2011}\)
XO
21 tháng 6 2019
\(\frac{5}{16}+\frac{5}{7}+0,4+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{25}{35}+\frac{2}{5}+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{25}{35}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{2}{3}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{5}{6}\)
= \(\frac{5}{16}+\frac{37}{30}\)
= \(\frac{371}{240}\)
12 tháng 4 2017
\(\frac{15}{37}\)x \(\left(\frac{38}{41}-\frac{74}{45}\right)-\frac{38}{41}\)x \(\left(\frac{15}{37}+\frac{82}{76}\right)\)
\(=\frac{15}{37}\)x \(\frac{38}{41}-\frac{15}{37}\)x \(\frac{74}{45}-\frac{38}{41}\)x \(\frac{15}{37}-\frac{38}{41}\)x \(\frac{82}{76}\)
\(=\frac{15}{37}\)x \(\frac{38}{41}-\frac{2}{3}-\frac{38}{41}\)x \(\frac{15}{37}-1\)
\(=\frac{15}{37}\)x \(\frac{38}{41}\)\(-\frac{38}{41}\)x \(\frac{15}{37}\)\(-\frac{2}{3}-1\)
\(=0-\frac{2}{3}-1=\frac{-5}{3}\)
HP
2 tháng 7 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2008}{6033}=\frac{1004}{6033}\)
2 tháng 7 2016
\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{2009x2011}\)
\(=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+....+\frac{1.2}{2009.2011.2}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2008}{6033}=\frac{2008}{12066}\)
7 tháng 5 2017
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
22 tháng 8 2016
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-1\frac{3}{4}\)
\(=2\frac{1}{4}\)
12 tháng 3 2022
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-\frac{7}{4}\)
\(=\frac{16}{4}-\frac{7}{4}\)
\(=\frac{9}{4}\)
22 tháng 8 2016
= \(6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}\)\(-1\frac{3}{4}\)= \(4-1\frac{3}{4}\)=\(\frac{9}{4}\)
22 tháng 8 2016
47/7 - (7/4 + 19/7) =47/7 -125/28 =gomen =tao xin lỗi chịu rùi = <3
22 tháng 2 2017
Ta có: B = \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}\)
=> B = \(\frac{6}{3.5}\)+ \(\frac{6}{5.7}\)+ \(\frac{6}{7.9}\)+ \(\frac{6}{9.11}\)
=>B =\(3.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
=> B = \(3.\left(\frac{1}{3}-\frac{1}{11}\right)\)
=> B = \(3.\frac{8}{33}\)
=> B = \(\frac{8}{11}\)
Vậy: B = \(\frac{8}{11}\)
\(\frac{2011x2010-1}{2009x2011+2010}=\frac{2011x\left(2009+1\right)-1}{2009x2011+2010}\)
\(=\frac{2011x2009+2011-1}{2009x2011+2010}=\frac{2011x2009+2010}{2009x2011+2010}=1\)