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a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 2006 ) + 2007 = 2007
\(\Rightarrow\)( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2006 + 2007 ) = 2007
\(\Rightarrow\)2007x + 2015028 = 2007
\(\Rightarrow\)2007x = 2007 - 2015028 = -2013021
\(\Rightarrow\)x = ( -2013021 ) : 2007 = -1003
Vậy x = -1003
b) 2000 + ( 199 + x ) + ( 198 + x ) + ... + ( x + 1 ) + x = 200
\(\Rightarrow\)( x + x + x + ... + x + x ) + ( 1 + 2 + ... + 198 + 199 + 2000 ) = 200
\(\Rightarrow\)200x + 2001000 = 200
\(\Rightarrow\)200x = 200 - 2001000 = -2000800
\(\Rightarrow\)x = ( -2000800 ) : 200 = -10004
Vậy x = -10004
a, x + ( x + 1 ) + ( x + 2 ) + ..... + ( x + 2006) + 2007 = 2007
x. 2007 + ( 1 + 2 + ..... + 2006 ) = 2007 - 2007
x. 2007 + 2013021 = 0
x. 2007 = 0 - 2013021
x.2007 = - 2013021
x = ( - 2013021 ) : 2007
x = - 1003
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}\)
\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\left(x-2010\right)\times\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
=> \(x-2010=0\)
\(x=2010\)
\(\dfrac{x-1}{2009}\)+\(\dfrac{x-2}{2008}\)=\(\dfrac{x-3}{2007}\)+\(\dfrac{x-4}{2006}\)
=>\(\dfrac{x-1}{2009}\)-1+\(\dfrac{x-2}{2008}\)+1=\(\dfrac{x-3}{2007}\)-1+\(\dfrac{x-4}{2006}\)-1
=>(x-2010)x(\(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)-\(\dfrac{1}{2006}\))=0
=>x-2010=0 (vì \(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)\(\dfrac{1}{2006}\)≠0)
=>x=2010
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009
=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2
=>1/2-1/(x+1)=2007/4018
=>1/(x+1)=1/2-2007/4018
=>1/x+1=1/2009
=>x+1=2009
=>x=2009-2008
=>x=1
vậy x=1
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
\(x-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x-x}{2007}=\frac{2006}{2007}\)
\(\frac{2006x}{2007}=\frac{2006}{2007}\Rightarrow2006x=2006\)
=>x=1
1/x+x+1+x+2+x+3+...+x+2006+2007=2007
------------------------------------------=2007-2007
------------------------------------------=0
x+x+x+...+x+1+2+3+...+2006=0
2007.x+(1+2+...+2006)=0
2007.x+(2006+1).[(2006-1)+1]:2=0
2007.x+2013021=0
2007.x=0-2013021
x=-2013021:2007
x=-1003
2/x+x+1+x+2+...+x+198=401-201-200-199
199.x+(1+2+...+198)=-199
199.x+(1+198).[(198-1)+1]:2=-199
199.x+19701=-199
199.x=-199-19701
x=-19900:199
x=-100
3/x+x+1+x+2+...+x+2008=2010-2010-2009
2009.x+(2008+1).[(2008-1)+1]:2=-2009
2009.x+2017036=-2009
2009.x=-2009-2017036
x=-2019045:2009
x=-1005