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NT
1
8 tháng 8 2018
\(\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}+\dfrac{x+5}{2007}+\dfrac{x+2007}{5}=-5\)
Ta có:
\(\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1+\dfrac{x+5}{2007}+1+\dfrac{x+2007}{5}+1=0\)
\(=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}+\dfrac{x+2012}{2007}+\dfrac{x+2012}{5}=0\)
\(=\left(x+2012\right)\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy \(x=-2012\)
Chúc bạn học tốt!
NT
1
AH
Akai Haruma
Giáo viên
13 tháng 9
Lời giải:
\(x=\frac{1}{2^{2009}}+\frac{2}{2^{2008}}+\frac{3}{2^{2007}}+....+\frac{2008}{2^2}+\frac{2009}{2}\)
\(2x = \frac{1}{2^{2008}}+\frac{2}{2^{2007}}+\frac{3}{2^{2006}}+...+\frac{2008}{2}+2009\)
\(\Rightarrow x=2x-x=2009-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2008}}-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)
\(\Rightarrow 2(2009-x)=1+\frac{1}{2}+....+\frac{1}{2^{2007}}+\frac{1}{2^{2008}}\)
\(\Rightarrow 2(2009-x)-(2009-x)=1-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=1-\frac{1}{2^{2009}}\\ \Rightarrow x=2009-(1-\frac{1}{2^{2009}})=2008+\frac{1}{2^{2009}}\)
PT
0
7 tháng 7 2017
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Rightarrow\left(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}\right)-\left(\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\right)=0\)
\(\Rightarrow\dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\)
\(\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\)
\(x+3=0\Rightarrow x=-3\)
7 tháng 7 2017
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\\ \dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\\ \left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x+3=0\Rightarrow x=-3\)
S
16 tháng 4 2017
b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
5 tháng 2 2020
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