So sánh: A =\(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
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23 tháng 6 2022
a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{202}{203}\)
b: \(=-4\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)
\(=-\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)
\(=\dfrac{-4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)
\(=\dfrac{-4}{3}\cdot\dfrac{504}{1009}=-\dfrac{672}{1009}\)
NT
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!
1 tháng 4 2023
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
24 tháng 7 2021
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
VT
6
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
\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
\(A=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
\(A=2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(A=2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(A=2.\dfrac{56}{305}\)
\(A=\dfrac{112}{305}\)
MV
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:
\(X=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(2X=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(2X=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2X-X=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)\(X=\dfrac{1}{2}-\dfrac{1}{2^{2016}}\)
\(Y=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\)
\(Y=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(Y=\dfrac{1}{5}-\dfrac{1}{61}=\dfrac{56}{305}\)
\(Z=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(Z=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{99.101}\)
\(Z=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(Z=\dfrac{7}{2}\left(1-\dfrac{1}{101}\right)\)
\(Z=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{700}{202}\)