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14 tháng 10 2017

=\(\frac{405^{30}}{75^{90}}\)

24 tháng 7 2016

\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+.......+\frac{61}{\left(30.31\right)^2}\)

   \(=\frac{1}{1^2.2^2}+\frac{1}{2^2.3^2}+....+\frac{1}{30^2.31^2}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{30}-\frac{1}{31}\)

    \(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{30}-\frac{1}{30}\right)-\frac{1}{31}\)

    \(=1-\frac{1}{31}\\ =\frac{31}{31}-\frac{1}{31}=\frac{30}{31}\)

no mình nha

    

3 tháng 9 2016

(\(\frac{121}{122}\)\(\frac{123}{125}\)) .(\(\frac{1995}{1996}\).\(\frac{17}{16}\)\(\frac{21}{25}\):\(\frac{16}{17}\)).(\(\frac{42}{30}\).\(\frac{75}{23}\)-\(\frac{19}{23}\).\(\frac{210}{38}\))

= (121/122 . 123/125) . ( 1995/1996 . 17/16 - 21/25 : 16/ 17) . 0

= 0

28 tháng 3 2017

\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{61}{\left(30.31\right)^2}\)

\(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{61}{30^2.31^2}\)

\(S=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{61}{900.961}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{900}-\frac{1}{961}\)

\(S=1-\frac{1}{961}\)

\(S=\frac{960}{961}\)

7 tháng 2 2017

\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)

\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(A=7.\frac{13}{28}\)

\(A=\frac{13}{4}\)

28 tháng 2 2020

                                                    Bài giải

a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)

\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)

b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)