HM
3
K
Khách
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TK
2
G
27 tháng 10 2017
Giải:
Ta có:
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-2=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}-2\)
\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}=\dfrac{x-3-2007}{2007}+\dfrac{x-4-2006}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\Leftrightarrow\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow x=2010\)
Vậy \(x=2010\).
Chúc bạn học tốt!
27 tháng 10 2017
| \(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\) | ||||
| \(\Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\) | ||||
chuyển vế ta có:
|
A
2
BH
17 tháng 3 2017
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
<=>\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
<=>\(\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010
OY
1
29 tháng 11 2020
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
12 tháng 8 2018
1,
x+1/2+x+1/3+x+1/4-x+1/5-x+1/6=0
(x+1)(1/2+1/3+1/4-1/5-1/6)=0
vì 1/2+1/3+1/4-1/5-1/6 khác 0
suy ra x+1=0 suy ra x=-1
BN
26 tháng 2 2018
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)
\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)
\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
20 tháng 12 2017
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) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
MC
6 tháng 3 2019
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)+\left(1+\frac{3}{2006}\right)\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
\(A=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
\(A=4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
Ta có: \(\left\{{}\begin{matrix}\frac{1}{2007}< \frac{1}{2006}\\\frac{1}{2008}< \frac{1}{2006}\\\frac{1}{2009}< \frac{1}{2006}\end{matrix}\right.\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}< \frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}=\frac{3}{2006}\)
\(\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}< 0\)
\(\Rightarrow4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)>4\)
hay \(A>4\)
\(\text{Vậy A>4}\)
NN
2
20 tháng 9 2021
\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)
20 tháng 9 2021
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Leftrightarrow x+3=0\)
hay x=-3
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
=>\(\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
=>x-2010=0
=>x=2010
(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