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27 tháng 4 2017

\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\)                                                                                                          \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)

27 tháng 4 2017

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(B=2.\frac{100}{101}=\frac{200}{101}\)

Ta có :

M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)

b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)

Ta có:

B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)

Vậy B= \(\dfrac{100}{101}\)

23 tháng 3 2022

 = \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)

 = \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

 = \(\dfrac{5}{2}.\dfrac{100}{101}\)

 = \(\dfrac{250}{101}\)

 

11 tháng 4 2018

2,

\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{4}\)

24 tháng 4 2017

cho minh xin yeu cau de bai

26 tháng 4 2017

trả hiểu yêu cầu đề bài là j cả

3 tháng 3 2017

2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)

= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

=1-\(\dfrac{1}{101}\)=...........

mk làm vậy thôi nha

thông cảm

leuleuyeu

2 tháng 3 2017

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)

=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)

tương tự

15 tháng 8 2023

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

1 tháng 7 2021

\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

1 tháng 7 2021

1. Tính 

\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)

\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)

2. Tính

\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)

\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)

\(=\dfrac{29}{35}+\dfrac{3}{4}\)

\(=\dfrac{116}{140}+\dfrac{105}{140}\)

\(=\dfrac{221}{140}\)

\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)

\(=\dfrac{9}{7}-\dfrac{55}{77}\)

\(=\dfrac{99}{77}-\dfrac{55}{77}\)

\(=\dfrac{44}{77}=\dfrac{4}{7}\)

\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)

\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)

\(=\dfrac{3}{5}\times\dfrac{9}{7}\)

\(=\dfrac{27}{35}\)

\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)

\(=\dfrac{14}{45}:\dfrac{3}{11}\)

\(=\dfrac{14}{45}\times\dfrac{11}{3}\)

\(=\dfrac{154}{135}\)

\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)

\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)

\(=\dfrac{41}{21}-\dfrac{1}{4}\)

\(=\dfrac{164}{84}-\dfrac{21}{84}\)

\(=\dfrac{143}{84}\)

\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)

\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)

\(=\dfrac{20}{27}\times\dfrac{2}{11}\)

\(=\dfrac{40}{297}\)

\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)

\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)

\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)

\(=\dfrac{32}{10}:\dfrac{2}{5}\)

\(=\dfrac{16}{5}\times\dfrac{5}{2}\)

\(=\dfrac{80}{10}=8\)

3 tháng 3 2023

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)

3 tháng 3 2023

`4/1.3+4/3.5+4/5.7+...+4/99.101`

`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`

`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`

`=2(1-1/101)`

`=2. 100/101`

`=200/101`

\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)

\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)

\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)

=168/101