CMR:
\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2008^3}< \dfrac{1}{4}\)
Những câu hỏi liên quan
1)Tính tỉ số \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
2) chứng minh: \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+..+\dfrac{19}{9^2.10^2}< 1\)
1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
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Giá trị biểu thức \(A=\dfrac{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+\dfrac{2005}{4}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2008}+\dfrac{1}{2009}}\) là \(A=........\)
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
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a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b, \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2008}\right).x=\dfrac{2009}{1}+\dfrac{2010}{2}+...+\dfrac{5016}{2008}-2008\)
c, \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{16}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2001}{2003}\)
GIÚP VỚI , MIK CẦN GẤP
Tính tỉ số \(\dfrac{A}{B}\) , biết:
\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\)
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
1. Cho Ndfrac{1}{31}+dfrac{1}{32}+...+dfrac{1}{60}CMR dfrac{3}{5} N dfrac{4}{5}2. Cho Mdfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+...+dfrac{29}{3^{29}}-dfrac{30}{3^{30}}CMR M dfrac{3}{16}3. Cho Qdfrac{2}{3}+dfrac{8}{9}+dfrac{26}{27}+...+dfrac{3^{2021}-1}{3^{2021}}CMR Qdfrac{4041}{2}
Đọc tiếp
1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
1.8,cho A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\).CMR:\(\dfrac{2}{5}< A< \dfrac{8}{9}\)
1.9,cho A=\(\dfrac{2}{3}+\dfrac{2}{5^2}+\dfrac{2}{7^2}+...+\dfrac{2}{2007^2}.CMR:A< \dfrac{1007}{2008}\)
Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
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26 tháng 2 2018 lúc 13:49
Tính tỉ số A/B, biết :
A\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\)
B\(=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)
\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)
\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
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Câu 1 Tìm x;y nguyên biết
1! +2!+3! +...+\(x!=y^2\)
Câu2
CMR tổng \(P=\dfrac{1}{3^2}-\dfrac{1}{3^4}+\dfrac{1}{3^6}-\dfrac{1}{3^8}+...+\dfrac{1}{3^{2006}}-\dfrac{1}{3^{2008}}< 0,1\)
Cố giúp trước tối ngày mi nha
Câu 1:
Câu hỏi của Trần Văn Thành - Toán lớp 6 - Học toán với OnlineMath
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Tính:
D =\(\dfrac{\dfrac{2010}{1}+\dfrac{2009}{2}+\dfrac{2008}{3}+...+\dfrac{1}{2010}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2011}}\)
Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)
= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))
= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)
= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))
Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)
=> D = 2011
cho mk 1 tick nha 
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