Cho M=(1-1/4).(1-1/9).(1-1/16)...(1-1/100)
Chứng tỏ:M<11/19
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LA
8 tháng 5 2017
đề bài bạn sai vì theo như quy luật thì :
A=\(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)
\(\dfrac{1}{4}>\dfrac{1}{3.2}\)
\(\dfrac{1}{9}>\dfrac{1}{3.4}\)
\(\dfrac{1}{16}>\dfrac{1}{4.5}\)
.
.
.
\(\dfrac{1}{81}>\dfrac{1}{9.10}\)
\(\dfrac{1}{100}>\dfrac{1}{10.11}\)
A > \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
A > \(\dfrac{1}{2}+\dfrac{1}{11}\) =\(\dfrac{13}{22}\)
mà \(\dfrac{13}{22}\)>\(\dfrac{65}{132}\) ; A>\(\dfrac{13}{22}\)
Vậy A>\(\dfrac{65}{132}\)
ND
1
S
4 tháng 5 2017
A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
= \(\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\right)\)
Ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\)
\(\frac{1}{4^2}>\frac{1}{4.5}\)
.........
\(\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{1}{4}+\frac{8}{33}=\frac{65}{132}\)
Vậy A > 65/132
NQ
0
MN
2
WH
6 tháng 5 2018
Ta có:
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Leftrightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{9}{22}\)
Ta lại có:
\(\frac{9}{22}=\frac{9.11}{22\cdot11}=\frac{99}{132}\)
Ta thấy: 99>65
\(\Rightarrow\frac{99}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\left(đpcm\right)\)
6 tháng 5 2018
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(A>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(A>\frac{33}{132}+\frac{44}{132}-\frac{12}{132}\)
\(A>\frac{65}{132}\)
TD
2
4 tháng 5 2017
A=1/2*2+1/3*3+1/4*4+...+1/10*10.
A>1/1*2+1/2*3+1/3*4+...+1/9*10.
A>1-1/2+1/2-1/3+...+1/9-1/10.
A>1-1/10.
A>9/10.
=>A>1/2.
Mà 1/2=66/132>65/132.
=>A>65/132.
Vậy A>65/132.
6 tháng 5 2017
A=1/2^2+1/3^2+1/4^2+......+1/9^2+1/10^2
=1/4+1/3×3+1/4×4+.....+1/9×9+1/10×10
=>A>1/4+(1/3×4+1/4×5+...+1/9×10+1/10×11)
=>A>1/4+(1/3-1/11)
=>A>1/4+8/33
=>A>65/132( đpcm)
TM
0
DT
0
NT
0
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{99}{100}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{9\cdot11}{10\cdot10}=\frac{\left(1\cdot2\cdot\cdot\cdot\cdot9\right)\cdot\left(3\cdot4\cdot\cdot\cdot10\cdot11\right)}{\left(2\cdot3\cdot\cdot\cdot\cdot9\cdot10\right)\left(2\cdot3\cdot\cdot\cdot\cdot\cdot10\right)}=\frac{11}{2.10}=\frac{11}{20}\)\(