Cho P = \(\dfrac{12}{1.4.7}+\dfrac{12}{7.4.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}.\) Chứng minh: P < \(\dfrac{1}{2}\).
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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)
DD
1
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
21 tháng 11 2017
\(\dfrac{12}{1.4.7}+\dfrac{12}{4.7.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}\)
\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+\dfrac{1}{7.10}-\dfrac{1}{10.13}+...+\dfrac{1}{54.57}-\dfrac{1}{57.60}\right)\)\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{57.60}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{57.60}\right)=\dfrac{1}{2}-\dfrac{1}{2.57.60}< \dfrac{1}{2}\left(đpcm\right)\)
20 tháng 7 2017
\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+...+\dfrac{1}{22.25.28}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1.4.7}+\dfrac{6}{4.7.10}+...+\dfrac{6}{22.25.28}\right)\)
\(=\dfrac{1}{6}\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+...+\dfrac{1}{22.25}-\dfrac{1}{25.28}\right)\)
\(=\dfrac{1}{6}\left(\dfrac{1}{4}-\dfrac{1}{25.28}\right)\)
\(=\dfrac{1}{24}-\dfrac{1}{6.25.28}\)
Vậy...
DH
20 tháng 7 2017
\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+\dfrac{1}{7.10.13}+...+\dfrac{1}{22.25.28}\)
\(\Rightarrow6S=\dfrac{6}{1.4.7}+\dfrac{6}{4.7.10}+\dfrac{6}{7.10.13}+...+\dfrac{6}{22.25.28}\)
\(\Rightarrow6S=\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+...+\dfrac{1}{22.25}-\dfrac{1}{25.28}\)
\(\Rightarrow6S=\dfrac{1}{1.4}-\dfrac{1}{25.28}\)
\(\Rightarrow6S=\dfrac{1}{4}-\dfrac{1}{700}=\dfrac{87}{350}\)
\(\Rightarrow S=\dfrac{29}{700}\)
Chúc bạn học tốt!!!
28 tháng 4 2021
Ta có: \(\dfrac{3}{10}>\dfrac{3}{15}\)
\(\dfrac{3}{11}>\dfrac{3}{15}\)
\(\dfrac{3}{12}>\dfrac{3}{15}\)
\(\dfrac{3}{13}>\dfrac{3}{15}\)
\(\dfrac{3}{14}>\dfrac{3}{15}\)
Do đó: \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}=1\)
hay 1<S(1)
Ta có: \(\dfrac{3}{11}< \dfrac{3}{10}\)
\(\dfrac{3}{12}< \dfrac{3}{10}\)
\(\dfrac{3}{13}< \dfrac{3}{10}\)
\(\dfrac{3}{14}< \dfrac{3}{10}\)
Do đó: \(\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{12}{10}\)
\(\Leftrightarrow S< \dfrac{15}{10}=\dfrac{3}{2}< 2\)(2)
Từ (1) và (2) suy ra 1<S<2(đpcm)
NT
Nguyễn thành Đạt
CTVHS
13 tháng 9 2023
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)
6 tháng 5 2021
bạn ơi cái câu <1 số hạng cuối cùng là j thế?
\(P=\dfrac{12}{1\cdot4\cdot7}+\dfrac{12}{4\cdot7\cdot10}+\dfrac{12}{7\cdot10\cdot13}+...+\dfrac{12}{54\cdot57\cdot60}\)
\(P=\dfrac{12}{6}\left(\dfrac{1}{1\cdot4}-\dfrac{1}{4\cdot7}+\dfrac{1}{4\cdot7}-\dfrac{1}{7\cdot10}+...+\dfrac{1}{54\cdot57}-\dfrac{1}{57\cdot60}\right)\)
\(P=2\left(\dfrac{1}{1\cdot4}-\dfrac{1}{57\cdot60}\right)\)
\(P=\dfrac{2}{4}-\dfrac{2}{57\cdot60}=\dfrac{1}{2}-\dfrac{1}{57\cdot30}\)
\(\Rightarrow P< \dfrac{1}{2}\)