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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!
| \(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\) | ||||
| \(\Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\) | ||||
chuyển vế ta có:
|
Giải:
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Leftrightarrow\dfrac{x+4}{2008}+\dfrac{x+3}{2009}+2=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}+2\)
\(\Leftrightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Leftrightarrow\dfrac{x+4+2008}{2008}+\dfrac{x+3+2009}{2009}=\dfrac{x+2+2010}{2010}+\dfrac{x+1+2011}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên \(x+2012=0\)
\(\Leftrightarrow x=0-2012\)
\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\).
Chúc bạn học tốt!
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên:
\(x+2012=0\Rightarrow x=-2012\)
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
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
<=>\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
<=>\(\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010
\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Leftrightarrow x+3=0\)
hay x=-3
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2011}+\dfrac{x-2}{2012}+\dfrac{x-3}{2009}-\dfrac{x-4}{2008}=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\)
\(\Leftrightarrow x-2012=0\Leftrightarrow x=2012\)
Vậy ...
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
=> \(\dfrac{x-1}{2011}-1+\dfrac{x-2}{2010}-1+\dfrac{x-3}{2009}-1=\dfrac{x-4}{2008}-1-2\)
=>\(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}=\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}\)
=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}=0\)=> x - 2012 ( \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\)) = 0
Vì \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\) \(\ge\) 0
=> x - 2012 = 0
=> x = 2012
\(x+2x+3x+...+2011x=2012.1013\)
\(\dfrac{2011\left(2011+1\right)}{2}x=2012.2013\)
\(x=2012.2013.\dfrac{2}{2011.2012}\)
\(x=\dfrac{4026}{2011}\)
a, H = \(2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Leftrightarrow\) 2H = \(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)
\(\Leftrightarrow\) 2H - H = \((2^{2011}-2^{2010}-2^{2009}-...-2^2-2)\) - \((2^{2010}-2^{2009}-2^{2008}-...-2-1)\)
\(\Leftrightarrow\) H = \(2^{2011}-2.2^{2010}+1\)
\(\Leftrightarrow\) H = \(2^{2011}-2^{2011}+1\)
\(\Leftrightarrow\) H = 1
Vậy H = 1
a)H=22010-22009-...-2-1
=>2H=2(22010-22009-...-2-1)
=>2H=22011-22010-...-22-2
=>2H-H=(22011-22010-...-22-2)-(22010-22009-...-2-1)
=>H=22011-1
a) Ta có:
\(-\dfrac{24}{35}< -\dfrac{24}{30}< -\dfrac{19}{30}\)
\(\Rightarrow x< y\)
b) Ta có:
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
Ta lại có:
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
=> Dễ dàng thấy A > B
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