Cho A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
Chứng tỏ rằng : A > \(\frac{65}{132}\)
Help meeee !!!
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Cho A=\(\frac{1}{4}\)+\(\frac{1}{9}\)+\(\frac{1}{16}\)+...+\(\frac{1}{81}\)+\(\frac{1}{100}\).Chứng tỏ rằng A>\(\frac{65}{132}\)
A=\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
A=\(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{9}+\frac{1}{10}-\frac{1}{10}\)
A= 0
=> A>\(\frac{65}{132}\)
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\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)Chứng minh:\(A>\frac{65}{132}\)
Ta có :
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{92}+\frac{1}{10^2}\)
Mà \(\frac{1}{3^2}>\frac{1}{3.4}\)
\(\frac{1}{4^2}>\frac{1}{4.5}\)
\(...\)
\(\frac{1}{9^2}>\frac{1}{9.10}\)
\(\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow A-\frac{1}{2^2}>\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow A-\frac{1}{2^2}>\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A-\frac{1}{2^2}>\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow A-\frac{1}{4}>\frac{8}{33}\)
\(\Rightarrow A>\frac{8}{33}+\frac{1}{4}\)
\(\Rightarrow A>\frac{65}{132}\left(dpcm\right)\)
CMR: A > \(\frac{65}{132}\)
A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)
\(\Rightarrow A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
\(A>1+0+0+0+...+0-\frac{1}{10}\)
\(A>1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)
mà \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\)
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Ta có : \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{3^2}+...+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\) \(\left(đpcm\right)\)
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Cho A= \(\frac{1}{4}\)+ \(\frac{1}{9}\)+ \(\frac{1}{16}\)+...+\(\frac{1}{81}\)+\(\frac{1}{100}\)
CMR : A >\(\frac{65}{132}\)
Help me
Ta có: \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
- Đến đây bn lấy \(\frac{9}{22}\) so sánh vs \(\frac{65}{132}\) là ra ĐPCM nhé :3
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\(Cho\)\(A\) = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+......+\frac{1}{81}+\frac{1}{100}\)\(.\)\(CMR\)\(A\)> \(\frac{65}{132}\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}+\frac{1}{10.10}\)
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.10}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
\(A>1+0+0+0+...+0-\frac{1}{10}\)
\(A>1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)
mà : \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\)
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Mình thấy trên mạng bạn Lê Mạnh Tiến Đat rất nổi vậy bạn có thể làm đc bài nầy ko???
Cho A=\(\frac{1}{4}\)+\(\frac{1}{9}\)+\(\frac{1}{16}\)+.....+\(\frac{1}{81}\)+\(\frac{1}{100}\)
CMR:A>\(\frac{65}{132}\)
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(A>\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2.2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2.2}+\frac{1}{3}-\frac{1}{11}\)
\(=\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\left(ĐPCM\right)\)
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mình định lm y hệt như lê mạnh tiến đạt
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Cho A=\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+......+\frac{1}{100}\)
Hãy so sánh A và \(\frac{65}{132}\)
Các bạn nhớ giải đầy đủ hộ mình nhé, bạn nào nhanh mình tick và kết bạn cho
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A< 1-\frac{1}{10}=\frac{9}{10}\)
\(=>A>\frac{65}{132}\)
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Bài 1: Chứng minh rằng:1)frac{1}{5} Afrac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{2007^2}2)Bfrac{1}{4}+frac{1}{9}+frac{1}{16}+...+frac{1}{81}+frac{1}{100}frac{65}{132}3)Cfrac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2} frac{3}{4}4)frac{1}{6} Dfrac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2}5)Efrac{1}{4^2}+frac{1}{6^2}+frac{1}{8^2}+...+frac{1}{100^2} frac{1}{4}Bài 2 : Cho Dfrac{12}{left(2cdot4right)^2}+frac{20}{left(4cdot6right)^2}+...+frac{388}{left(96cdot98right)^2}+frac...
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Bài 1: Chứng minh rằng:
1)\(\frac{1}{5}< A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}\)
2)\(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}>\frac{65}{132}\)
3)\(C=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{3}{4}\)
4)\(\frac{1}{6}< D=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
5)\(E=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Bài 2 : Cho \(D=\frac{12}{\left(2\cdot4\right)^2}+\frac{20}{\left(4\cdot6\right)^2}+...+\frac{388}{\left(96\cdot98\right)^2}+\frac{396}{\left(98\cdot100\right)^2}\)
Hãy so sánh\(D\) với \(\frac{1}{4}\)
Cảm ơn các bạn nhiều!
Chứng tỏ rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)