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28 tháng 9 2018

KO HIỂU J CẢ

28 tháng 9 2018

\(\frac{E.I.G.H.T}{F.O.U.R}\) =    \(\frac{8}{4}\) =   2 

K NHA 

18 tháng 2 2017

\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{899}{900}\)

\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times...\times\frac{29.31}{30.30}\)

\(=\frac{\left(1\times2\times3\times...\times29\right)\left(3\times4\times5\times...\times31\right)}{\left(2\times3\times4\times...\times30\right)\left(2\times3\times4\times...\times30\right)}\)

\(=\frac{1\times2\times3\times...\times29}{2\times3\times4\times...\times30}.\frac{3\times4\times5\times...\times31}{2\times3\times4\times...\times30}\)

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

\(=\frac{31}{60}\)

19 tháng 2 2017

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{29.31}{30.30}\\ =\frac{1.2.3.4....29}{2.3.4...30}.\frac{3.4.5...31}{2.3.4...30}\\ =\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)

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17 tháng 4 2019

G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)                         

=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)

=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)

=> G = \(\frac{2.50}{1.51}\)

=> G = \(\frac{100}{51}\)

17 tháng 4 2019

公关稿黄继线长旧款您

16 tháng 3 2022

\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)

\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)

\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)

\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)

a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)

\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)

\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)

\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)

b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)

c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)

\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)

\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)

d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)

\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)

e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)

\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)

g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)

\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)

14 tháng 7 2020

thank you,very well

2 tháng 4 2017

B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)

B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)

B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)

B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)

B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)

B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)

B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)

B=\(\dfrac{17}{25}\)

3 tháng 4 2016

a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)

=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)

=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)

=\(-\frac{11}{4}\cdot\frac{8}{33}\)

=\(-\frac{2}{3}\)

b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)

=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)

=\(\frac{-1}{11}\cdot55=-5\)

c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)

=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)

=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)

=\(-\frac{2}{3}\cdot\frac{7}{5}\)

=\(\frac{-14}{15}\)

d) chưa nghĩ ra nhé

e) bạn chép sai đề bài rồi

mk mới kiểm tra 45 phút nên biết

đề bài nè

\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)

=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)

=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)

=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)

=\(\frac{1.31}{2.30}\)

=\(\frac{31}{60}\)

3 tháng 4 2016

a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong                                                                         b)bn tu lam nhe                                                                                                                                             c)dat thua so chung                                                                                                                                       d)tinh trong ngoac ra rui nhan vs                                                                                                                       e) mk bo tay 

26 tháng 4 2017

kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi

26 tháng 4 2017

trời ơi bài dễ thế này tự làm đi còn hỏi