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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
Lời gải:
a. Số số hạng:
$(2007-x):1+x=2008-x$
Suy ra:
$x+(x+1)+(x+2)+....+2006+2007=2007$
$\frac{(x+2007)(2008-x)}{2}=2007$
$(x+2007)(2008-x)=4014=$
$\Rightarrow x=2007$ hoặc $x=-2006$
b.
Số số hạng: $(2000-x):1+1=2001-x$
Suy ra:
$2000+1999+...+(x+1)+x=2000$
$\frac{(2000+x)(2001-x)}{2}=2000$
$(2000+x)(2001-x)=4000$
$\Rightarrow x=2000$ hoặc $x=-1999$
\(x-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x-x}{2007}=\frac{2006}{2007}\)
\(\frac{2006x}{2007}=\frac{2006}{2007}\Rightarrow2006x=2006\)
=>x=1
Để |x+2007|+|x+2006|=1
Ta có |x+2007|>|x+2006|. Suy ra x+2007=1 và x+2006=0 suy ra x=-2006
Ai k mình k lại gấp 10
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}\)
\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\left(x-2010\right)\times\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
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
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