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14 tháng 11 2017

\(\frac{5}{66}\)    nha

14 tháng 11 2017

\(B=\frac{5}{11.13}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(\Rightarrow B=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(\Rightarrow B=\frac{1}{11}-\frac{1}{66}\)

\(\Rightarrow B=\frac{5}{66}\)

14 tháng 11 2017

\(B=\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{61.66}\)

\(B=\dfrac{5}{5}\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)

\(B=\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\)

\(B=\dfrac{1}{11}-\dfrac{1}{66}\)

\(B=\dfrac{6}{66}-\dfrac{1}{66}=\dfrac{5}{66}\)

22 tháng 7 2023

D=\(\dfrac{4}{11\cdot16}\)+\(\dfrac{4}{16\cdot21}\)+...+\(\dfrac{4}{61\cdot66}\)

D=\(\dfrac{4}{5}\)(\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+\(\dfrac{1}{16}\)-\(\dfrac{1}{21}\)+...+\(\dfrac{1}{61}\)-\(\dfrac{1}{66}\))

D=\(\dfrac{4}{5}\)(\(\dfrac{1}{11}\)-\(\dfrac{1}{66}\))

D=\(\dfrac{4}{5}\)x\(\dfrac{5}{66}\)=\(\dfrac{2}{33}\)

10 tháng 9 2016

a)

\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)

\(\Rightarrow A=\frac{17}{35}\)

b)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)

c)

\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow C=1-1-\frac{1}{25}\)

\(\Rightarrow C=\frac{1}{25}\)

 

20 tháng 9 2023

Đặt:

\(A=\dfrac{7}{11\cdot16}+\dfrac{7}{16\cdot21}+\dfrac{7}{21\cdot26}+...+\dfrac{7}{61\cdot66}\)

\(\dfrac{5}{7}A=\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{61\cdot66}\)

\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\)

\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{66}=\dfrac{6}{66}-\dfrac{1}{66}=\dfrac{5}{66}\)

\(A=\dfrac{5}{66}\cdot\dfrac{7}{5}=\dfrac{7}{66}\)

23 tháng 10 2016

a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)

\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(=\frac{1}{5}+\frac{2}{7}\)

\(=\frac{7}{35}+\frac{10}{35}\)

\(=\frac{17}{35}\)

Vậy \(A=\frac{17}{35}\)

b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)

\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)

\(=5.\frac{50}{671}\)

\(=\frac{250}{671}\)

Vậy \(B=\frac{250}{671}\)

13 tháng 8 2018

P/s: làm từng phần một

1.

\(2A=2^2+2^3+...+2^{101}\)

\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(A=2^{101}-2\)

13 tháng 8 2018

2.

\(\frac{A}{2}=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\)

\(\frac{A}{2}=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)

\(\frac{A}{2}=\frac{1}{5}-\frac{1}{61}\)

\(\frac{A}{2}=\frac{56}{305}\)

\(A=\frac{112}{305}\)

26 tháng 10 2015

đang bận làm để thông cảm nha có j kiếm lại chất xám mình giải cho

12 tháng 8 2016

bạn ơi hình như đề sai ở chỗ cuối cùng kia kìa chỗ đó có phải : x . x ( 1 + 5 ) 

Đúng ko bạn ?????

12 tháng 8 2016

Sai đề

21 tháng 11 2018

a) \(A=\dfrac{5^2}{11.16}+\dfrac{5^2}{16.21}+\dfrac{5^2}{21.26}+...+\dfrac{5^2}{56.61}\)

\(A=5^2.\left(\dfrac{1}{11.16}+\dfrac{1}{16.21}+\dfrac{1}{21.26}+...+\dfrac{1}{56.61}\right)\)

\(A=\left(5^2:5\right).\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+...+\dfrac{5}{56.61}\right)\)

\(A=5.\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+...+\dfrac{1}{56}-\dfrac{1}{61}\right)\)

\(A=5.\left(\dfrac{1}{11}-\dfrac{1}{61}\right)\)

\(A=5.\dfrac{50}{671}\)

\(Á=\dfrac{250}{671}\)

22 tháng 11 2022

b: \(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)

\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=-2\cdot\dfrac{49}{50}=-\dfrac{49}{25}\)