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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B= \(\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) \)
=\(\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) \)
=\(\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )\)
=\(\frac{2}{2009}-1 -\frac{1}{2010} \)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-66.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(=-66.\dfrac{17}{66}+124.\left(-37-63\right)\)
\(=-17+124.\left(-100\right)=-17-12400\)
\(=-12417\)
Tính :
-66 . ( \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11})\)+ 124 . (-37) + 63 . (-124)
= -66 . (\(\dfrac{33}{66}-\dfrac{22}{66}+\dfrac{6}{66}\))+ (-4588) + (-7812)
= -66 . \(\dfrac{17}{66}\)+ (-12400)
= -17 + (-12400)
= -12417
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)
\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)
\(=\dfrac{1}{n+1}\)
2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)
=-12405
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)
\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)