2008.(\(\dfrac{1}{2007}\)-\(\dfrac{2009}{1004}\))-2009.(\(\dfrac{1}{2007}\)-2)
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Những câu hỏi liên quan
29 tháng 5 2022
a: \(=\dfrac{2008}{2007}-2009\cdot2-\dfrac{2009}{2007}+2009\cdot2\)
=-1/2007
b: \(=\dfrac{5^5\cdot5^3\cdot2^6-5^4\cdot5^3\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^8\cdot2^6-5^7\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^7\cdot2^6\left(5-1+2^4\right)}{5^6\cdot2^{10}}=\dfrac{5}{16}\cdot\dfrac{20}{1}=\dfrac{100}{16}=\dfrac{25}{4}\)
4 tháng 11 2018
1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)
=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)
=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)
=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)
=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)
=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)
=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)
=\(\dfrac{1}{2007}.\left(-1\right)+0\)
=\(\dfrac{-1}{2007}\)
4 tháng 11 2018
2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)
=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)
=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)
=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)
=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)
bí rồi
PT
2
29 tháng 11 2020
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
LG
12 tháng 7 2017
Đặt \(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}\) và \(B=\dfrac{2009^{2007}+1}{2009^{2008}+1}\)
Ta có:
\(2009A=\dfrac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\dfrac{2009^{2009}+2009}{2009^{2009}+1}\)
\(=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}\)
\(=1+\dfrac{1}{2009^{2009}+1}\)
\(2009B=\dfrac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\dfrac{2009^{2008}+2009}{2009^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2008}{2009^{2008}+1}=\dfrac{2008^{2008}+1}{2009^{2008}+1}+\dfrac{2008}{2009^{2008}+1}\)
\(=1+\dfrac{2008}{2009^{2008}+1}\)
Vì \(1+\dfrac{2008}{2009^{2009}+1}< 1+\dfrac{2008}{2009^{2008}+1}\)
Nên \(10A< 10B\) \(\Rightarrow A< B\)
Vậy \(\dfrac{2009^{2008}+1}{2009^{2009}+1}< \dfrac{2009^{2007}+1}{2009^{2008}+1}\)
~ Học tốt ~
12 tháng 7 2017
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}< 1\)
\(\Rightarrow A< \dfrac{2009^{2008}+1+2008}{2009^{2009}+1+2008}\Rightarrow A< \dfrac{2009^{2008}+2009}{2009^{2009}+2009}\Rightarrow A< \dfrac{2009\left(2009^{2007}+1\right)}{2009\left(2009^{2008}+1\right)}\Rightarrow A< \dfrac{2009^{2007}+1}{2009^{2008}+1}=B\)\(\Rightarrow A< B\)
HM
3
23 tháng 10 2023
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
=>x-2010=0
=>x=2010
KV
23 tháng 10 2023
(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006
(x - 1)/2009 - 1 + (x - 2)/2008 - 1 = (x - 3)/2007 - 1 + (x - 4)/2006 - 1
(x - 2010)/2009 + (x - 2010)/2008 = (x - 2010)/2007 + (x - 2010)/2006
(x - 2010)/2009 + (x - 2010)/2008 - (x - 2010)/2007 - (x - 2010)/2006 = 0
(x - 2010).(1/2009 + 1/2008 - 1/2007 - 1/2006) = 0
x - 2010 = 0
x = 2010
19 tháng 11 2017
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
LL
30 tháng 4 2017
Ta có:
\(2007A=\dfrac{2007^{2009}+2007}{2007^{2009}+1}=1+\dfrac{2006}{2007^{2009}+1}\)\(2007B=\dfrac{2007^{2010}+10}{2007^{2010}+1}=1+\dfrac{9}{2007^{2010}+1}\)Vì \(\dfrac{2007}{2007^{2009}+1}>\dfrac{2007}{2007^{2010}+1}\)
=>2007A > 2007B
Do đó A>B
Vậy A>B
30 tháng 4 2017
Ta có : \(B\) = \(\dfrac{2007^{2009}+1}{2007^{2010}+1}\) \(< 1\) \(\Rightarrow\dfrac{2007^{2009}+1}{2007^{2010}+1}< \dfrac{2007^{2009}+1+2006}{2007^{2010}+1+2006}\) \(=\dfrac{2007^{2009}+2007}{2007^{2010}+2007}\)
\(=\dfrac{2007\left(2007^{2008}+1\right)}{2007\left(2007^{2009}+1\right)}\) \(=\dfrac{2007^{2008}+1}{2007^{2009}+1}=A\)
Vậy \(A>B\)
BN
26 tháng 2 2018
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)
\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)
\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
\(=\dfrac{2008}{2007}-2\cdot2009-\dfrac{2009}{2007}+2009\cdot2\)
=-1/2007