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28 tháng 4 2018
a/ Tinh giá trị:
\(D=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{10}\right)\) \(\Leftrightarrow D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{7}{8}.\frac{8}{9}.\frac{9}{10}=\frac{1}{10}\)
b/ Chứng minh:
\(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
- Với mọi số tự nhiên n khác không thì luôn có: \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\) Do đó:
\(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}=\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{101}\right)< \frac{1}{2}\) Vậy \(E< \frac{1}{2}\)
c/ Chứng minh : \(F=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\)
\(F=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)>\frac{50}{150}+\frac{50}{200}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
Vậy: \(F>\frac{7}{12}\) .
HN
1
5 tháng 4 2016
Các bạn nào giúp mình trong tối nay 4 \ 5 \2016 sẽ được nha ...
DL
0
HN
1
NT
2
7 tháng 5 2016
Xét: 1 / 4 > 1 / 16 ; 1 / 5 > 1 / 16
=) 1 / 4 + 1 / 5 + 1 / 6 + ... + 1 / 19 > 16 . 1 / 16 = 16 /16 = 1
=) B > 1
Vậy B > 1
18 tháng 4 2017
B= 1/4+(1/5+1/6+...+1/9)+(1/10+1/11+...+1/19)
Vì 1/5+1/6+...+1/9 > 1/9+1/9+...+1/9 nên 1/5+1/6+...+1/9 > 5/9 >1/2
Vì 1/10+1/11+...+1/19 > 1/19+1/19+...+1/19 nên 1/10+1/11+...+1/19 > 10/19 >1/2
Suy ra: B > 1/4+1/2+1/2 > 1
NP
1
TT
1
21 tháng 4 2016
B = 1/4 + 1/5 + 1/6 +....+1/19
> 1/4 + ( 1/20 + 1/20 +.....+1/20) ( 15 p/s 1/20) = 1/4 + 3/4 = 1
=> B > 1
Vậy B > 1
TK
11
29 tháng 4 2017
Ta có :
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)
Vì \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)
\(\Rightarrow B>1\)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(\Rightarrow B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\right)\)Do \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>4.\frac{1}{8}=\frac{1}{2}\left(1\right)\)
\(\Rightarrow\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}>4.\frac{1}{12}=\frac{1}{3}\left(2\right)\)
\(\Rightarrow\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}>4.\frac{1}{16}=\frac{1}{4}\left(3\right)\)
\(\Rightarrow\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}>4.\frac{1}{20}=\frac{1}{5}\left(4\right)\)
Từ (1) , ( 2 ) , ( 3 ) và ( 4 ) suy ra :
\(B>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(B>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{5}\)
\(B>\frac{1}{2}+\frac{1}{2}+\frac{1}{5}\)
\(B>1+\frac{1}{5}\Rightarrow B>1\)
Vậy : \(B>1\)
Ta có :
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\)
\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\)
\(\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\right)\)
\(B>\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\left(\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\right)+\)
\(\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\right)\)
\(B>\frac{4}{8}+\frac{4}{12}+\frac{4}{16}+\frac{4}{20}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{77}{60}>\frac{60}{60}=1\)
\(\Rightarrow\)\(B>1\)
Vậy \(B>1\)
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