Bài 1: Tìm x biết:
1/ x + (x + 1) + (x + 2) + (x + 3) +.....+ (x + 2006) + 2007 = 2007
2/ x + (x + 1) + (x + 2) +...+ 199 + 200 + 201 = 401
3/ x + (x + 1) + (x + 2) +....+ 2009 + 2010 = 2010
Bn nào nhanh mik tick nha!!!
K
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13 tháng 1 2018
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
13 tháng 1 2018
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
LM
0
T
1
3 tháng 1 2021
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)
\(\Leftrightarrow x=2010\)
NT
1
11 tháng 2 2017
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
\(\Rightarrow x-2010=0\Rightarrow x=2010\)
K
1
4 tháng 5 2019
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)
<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)
<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0
<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0
<=> x+2011=0
<=> x=-2011
Vậy pt có nghiệm là x=-2011
OY
1
NT
1
DN
1 tháng 8 2015
(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1]
=(x-2011)/2007+(x-2011)/2008=(x-2011)/...
=(x-2011)(1/2007+1/2008-1/2009-1/2010)...
suy ra x=2010
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