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10 tháng 5 2018

A=3/4.(1/5.7+1/7.9+....+1/59.61)

A=3/4.(1/5-1/7+1/7-1/9+...+1/59-1/61)

A=3/4.(1/5-1/61)

A=3/4.56/305

A=42/305

mình làm cho bạn phần A thôi nhé còn phần B mình chưa nghĩ ra cách làm ahihi!

12 tháng 8 2017

a,

\(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\\ =1\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\cdot\dfrac{1}{2^2}+\left(2-1\right)\cdot\dfrac{1}{2^3}+...+\left(2-1\right)\cdot\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}-\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}}{2^{2006}}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}-1}{2^{2006}}\)

b,

\(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\\ =\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\\ =\dfrac{1}{5}-\dfrac{1}{61}\\ =\dfrac{56}{305}\)

c,

\(\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{9999}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\dfrac{100}{101}\\ =\dfrac{350}{101}\)

12 tháng 8 2017

Đặt:

\(X=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

\(2X=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)

\(2X=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(2X-X=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)\(X=\dfrac{1}{2}-\dfrac{1}{2^{2016}}\)

\(Y=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\)

\(Y=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

\(Y=\dfrac{1}{5}-\dfrac{1}{61}=\dfrac{56}{305}\)

\(Z=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)

\(Z=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{99.101}\)

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

\(Z=\dfrac{7}{2}\left(1-\dfrac{1}{101}\right)\)

\(Z=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{700}{202}\)

13 tháng 8 2017

a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)

\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)

\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)

b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)

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

\(A=\dfrac{1}{5}-\dfrac{1}{61}\)

\(A=\dfrac{56}{305}\)

c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)

\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)

\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)

\(A=\dfrac{7}{2}.\dfrac{100}{101}\)

\(A=\dfrac{256}{101}\)

17 tháng 4 2017

b,=1/5-1/7+1/7-1/9+...+1/59-1/61

=1/5-1/61

=54/115

11 tháng 5 2022

ơi

11 tháng 5 2022

30 tháng 3 2017

2)

a) M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)

30 tháng 3 2017

1) Quãng đường AB là :
\(36\dfrac{1}{4}\)km/h . 3.2h = 116 ( km )
Thời gian người ấy đi từ A đến B lúc về là :
116 : 40 = 2.9 ( giờ )
Đ/S : 2.9 giờ