Tìm các số nguyên x biết:
a,x+(x+1)+(x+2)+...+2006+2007=2007
b,2000+1999+...+x+1+x=2000
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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
DT
1
PU
3
20 tháng 9 2019
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+1010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)=\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\right)\)
\(\Rightarrow x+2010=0\) vì \(0< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}< \frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\)
\(\Rightarrow x=-2010\)
F
20 tháng 9 2019
Bài giải
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-(\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998})=0\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\) nên \(x+2010=0\)
\(x=0-2010=-2010\)
KB
1
8 tháng 7 2017
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)
\(\Rightarrow\left(\dfrac{x+1}{2009}+1\right)+\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)=\left(\dfrac{x+10}{2000}+1\right)+\left(\dfrac{x+11}{1999}+1\right)+\left(\dfrac{x+12}{1998}+1\right)\)
\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}=\dfrac{x+2010}{2000}+\dfrac{x+2010}{1999}+\dfrac{x+2010}{1998}\)\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2000}-\dfrac{x+2010}{1999}-\dfrac{x+2010}{1998}=0\)\(\Rightarrow\left(x+2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2007}-\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}\right)=0\)\(\Rightarrow x+2010=0\Rightarrow x=-2010\)
NH
10 tháng 7 2019
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
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 $x=-1999$
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