\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)

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11 tháng 3 2017

Ta có :

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)

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

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

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

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

\(\Rightarrow A=\dfrac{112}{305}\)

Chúc bn học tốt!!

11 tháng 3 2017

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)

\(A=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

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

\(A=2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(A=2.\dfrac{56}{305}\)

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

11 tháng 3 2017

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}..........+\dfrac{4}{59.61}\)

\(\dfrac{1}{2}A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+.........+\dfrac{2}{59.61}\)

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

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

\(\dfrac{1}{2}A=\dfrac{56}{305}\)

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

11 tháng 3 2017

A=4(\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{59.61}\))

=4(\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{59}\)-\(\dfrac{1}{61}\))

=4(\(\dfrac{1}{5}\)-\(\dfrac{1}{61}\))

=4.\(\dfrac{56}{305}\)

=\(\dfrac{224}{305}\)

11 tháng 3 2017

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)

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

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

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

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

\(\Rightarrow A=\dfrac{112}{305}\)

11 tháng 3 2017

Được

11 tháng 3 2017

2 * {2/5*7 + 2/7*9 +............+2/59*61}

2 * {1/5 - 1/7 + 1/7 - 1/9 +.........+1/59 - 1/61}

2 * {1/5 - 1/61}

2 * 56/305 =112/305

chúc bạn học tốt

17 tháng 4 2017

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

=1/5-1/61

=54/115

20 tháng 3 2017

\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)

\(=3\left(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{59\cdot61}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) \(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

20 tháng 3 2017

A=\(\dfrac{3}{5.7}\)+\(\dfrac{3}{7.9}\)+...+\(\dfrac{3}{59.61}\)

A=\(\dfrac{3}{2}\)(\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+...+\(\dfrac{2}{59.61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+..+\(\dfrac{1}{59}\)-\(\dfrac{1}{61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{61-5}{5.61}\))

A=\(\dfrac{3}{2}\).\(\dfrac{56}{305}\)

A=\(\dfrac{84}{305}\)

leuleuok

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!

2 tháng 5 2017

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)

\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)

2 tháng 5 2017

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

= \(\dfrac{1}{3}-\dfrac{1}{61}\)

= \(\dfrac{58}{183}\)

14 tháng 3 2017

\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

14 tháng 3 2017

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)

\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)

\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}.\dfrac{56}{305}\)

\(=\dfrac{84}{305}\)

27 tháng 3 2017

Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?

27 tháng 3 2017

A=1-\(\dfrac{4}{5.7}-\dfrac{4}{7.9}-\dfrac{4}{9.11}...-\dfrac{4}{59.61}\)

A=\(1-\left(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\right)\)

Đặt B=\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\)

B=\(2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\right)\)
B=\(2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
B=\(2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=2.\dfrac{56}{305}\)
B=\(\dfrac{112}{305}\)
\(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)

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

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

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

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

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

13 tháng 1

Ta có: \(A=\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}\)

\(=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}\right)\)

\(=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}\right)\)

\(=3\left(\frac12-\frac{1}{11}\right)=3\cdot\frac{9}{22}=\frac{27}{22}\) >1

Ta có: \(B=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\cdots+\frac{4}{59\cdot61}\)

\(=2\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\cdots+\frac{2}{59\cdot61}\right)\)

\(=2\left(\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac15-\frac{1}{61}\right)=2\cdot\frac{61-5}{305}=2\cdot\frac{56}{305}=\frac{112}{305}<1\)

Ta có: A>1

B<1

Do đó: A>B