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\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+.....+\dfrac{1}{20}\)\(< \dfrac{1}{2}\)
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9 tháng 8 2018
Ta có : A = 1/10 + 1/12 + 1/14 + ... + 1/20 > 1/20 + 1/20 + ... + 1/20 . ( 10 số hạng ) = 1/20 * 10 . = 1/2 . Do đó A > 1/2 . Vậy bài toán được chứng minh .
20 tháng 3 2017
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
VD
4 tháng 5 2018
\(\dfrac{5}{2}A=\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{9-4}{4.9}+\dfrac{14-9}{9.14}+\dfrac{19-14}{14.19}+...+\dfrac{509-504}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{509}\)
\(A=\left(\dfrac{1}{4}-\dfrac{1}{509}\right).\dfrac{2}{5}\)
\(A=\dfrac{1}{10}-\dfrac{2}{2545}< \dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{1}{10}\)(đpcm)
Chúc bạn học tốt!
LB
4 tháng 5 2018
Ta có:
A=\(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)
A=2.(\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{504.509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{509}{2036}-\dfrac{4}{2036}\))
A=\(\dfrac{2}{5}\).\(\dfrac{505}{2036}\)
A=\(\dfrac{101}{1018}\)
AM
8 tháng 6 2021
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
8 tháng 6 2021
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
29 tháng 4 2017
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
30 tháng 4 2017
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 9 2023
a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\)\(x\) = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)
\(\dfrac{2}{5}\) \(x\) = - \(\dfrac{3}{20}\)
\(x\) = - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)
\(x\) = - \(\dfrac{3}{8}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 9 2023
b, \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)
\(x\) = \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))
\(x\) = - \(\dfrac{15}{16}\)
27 tháng 4 2023
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
23 tháng 6 2018
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
Ta có :
\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}\left(6PS\right)\)
Mà\(\dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}=6.\dfrac{1}{12}=\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{2}\\ \dfrac{1}{10}>\dfrac{1}{12}\\ \dfrac{1}{12}=\dfrac{1}{12}\\ ...\\ \dfrac{1}{20}< \dfrac{1}{12}\)
⇒Cộng 2 vế, ta có:
\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{6}{12}=\dfrac{1}{2}\)
Vậy A<\(\dfrac{1}{2}\)