Tính
\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+\dfrac{1}{7.10.13}+...+\dfrac{1}{22.25.28}\)
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LM
7 tháng 3 2017
\(P=\dfrac{12}{1\cdot4\cdot7}+\dfrac{12}{4\cdot7\cdot10}+\dfrac{12}{7\cdot10\cdot13}+...+\dfrac{12}{54\cdot57\cdot60}\)
\(P=\dfrac{12}{6}\left(\dfrac{1}{1\cdot4}-\dfrac{1}{4\cdot7}+\dfrac{1}{4\cdot7}-\dfrac{1}{7\cdot10}+...+\dfrac{1}{54\cdot57}-\dfrac{1}{57\cdot60}\right)\)
\(P=2\left(\dfrac{1}{1\cdot4}-\dfrac{1}{57\cdot60}\right)\)
\(P=\dfrac{2}{4}-\dfrac{2}{57\cdot60}=\dfrac{1}{2}-\dfrac{1}{57\cdot30}\)
\(\Rightarrow P< \dfrac{1}{2}\)
DD
1
18 tháng 2 2017
Gọi biểu thức là A, ta có:
A = \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}=2\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+\frac{6}{7.10.13}+...+\frac{6}{54.57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-\frac{1}{10.13}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{57.60}\right)=2\left(\frac{427}{1710}\right)=\frac{427}{855}< \frac{427}{854}=\frac{1}{2}\)
Vậy A < \(\frac{1}{2}\)(điều cần chứng minh)
26 tháng 7 2019
Câu hỏi của thục hà - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
KN
26 tháng 7 2019
Đề sai hả
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{57.60}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}.2=\frac{1}{2}\)
3 tháng 4 2019
1
B= 12/1.4.7 + 12/4.7.10 + 12/7.10.13 + ... + 12/54.57.60
=> 1/2B= 6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60
=> 1/2B = 1/1.4 - 1/4.7 +1/4.7 - 1/7.10 +1/7.10 - 1/10.13 + ... + 1/54.57 - 1/57.60
=> 1/2B =1/1.4 - 1/57.60
=> 1/2B = 1/4 - 1/3420
=> 1/2B = 427/1710
=> B = 427/1710 . 2
=> B = 427/855
3 tháng 4 2019
2
A= 1+ 1/22 + 1/32 +...+1/1002
=1+ 1/2.2 + 1/3.3 +...+ 1/100.100
=> A< 1+ 1/1.2 + 1/2.3 +...+ 1/99.100
= 1+ 1 - 1/2 +1/2 - 1/3 +...+1/99 - 1/100
= 2- 1/100 < 2
Vậy A < 2
B
2
13 tháng 5 2019
Ta có \(A=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-...+\frac{1}{16.19}-\frac{1}{19.22}\)
\(=\frac{1}{4}-\frac{1}{418}=\frac{207}{836}\)
13 tháng 5 2019
\(A=\frac{6}{1\cdot4\cdot7}+\frac{6}{4\cdot7\cdot10}+\frac{6}{7\cdot10\cdot13}+...+\frac{6}{16\cdot19\cdot22}\)
\(A=\frac{1}{1\cdot4}-\frac{1}{4\cdot7}+\frac{1}{4\cdot7}-\frac{1}{7\cdot10}+...+\frac{1}{16\cdot19}-\frac{1}{19\cdot22}\)
\(A=\frac{1}{4}-\frac{1}{19\cdot22}=\frac{207}{836}\)
12 tháng 12 2021
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)
\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+...+\dfrac{1}{22.25.28}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1.4.7}+\dfrac{6}{4.7.10}+...+\dfrac{6}{22.25.28}\right)\)
\(=\dfrac{1}{6}\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+...+\dfrac{1}{22.25}-\dfrac{1}{25.28}\right)\)
\(=\dfrac{1}{6}\left(\dfrac{1}{4}-\dfrac{1}{25.28}\right)\)
\(=\dfrac{1}{24}-\dfrac{1}{6.25.28}\)
Vậy...
\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+\dfrac{1}{7.10.13}+...+\dfrac{1}{22.25.28}\)
\(\Rightarrow6S=\dfrac{6}{1.4.7}+\dfrac{6}{4.7.10}+\dfrac{6}{7.10.13}+...+\dfrac{6}{22.25.28}\)
\(\Rightarrow6S=\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+...+\dfrac{1}{22.25}-\dfrac{1}{25.28}\)
\(\Rightarrow6S=\dfrac{1}{1.4}-\dfrac{1}{25.28}\)
\(\Rightarrow6S=\dfrac{1}{4}-\dfrac{1}{700}=\dfrac{87}{350}\)
\(\Rightarrow S=\dfrac{29}{700}\)
Chúc bạn học tốt!!!