Tìm số hữu tỉ x,biết : \frac{x+6}{2010}+\frac{x+5}{2009}=\frac{x+4}{2008}+\frac{x+4}{2007}
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6 tháng 2 2018
Ta có :
\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)
\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)
\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)
Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)
\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài
Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)
WJ
1
NQ
30 tháng 7 2015
\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)
\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)
\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)
\(\Leftrightarrow\)\(x+2011=0\)
\(\Rightarrow x=0-2011\)
\(\Rightarrow x=-2011\)
7 tháng 1 2016
(x+1)/2011+1+(x+2)/2010+1+(x+3)/2009+1-((x+4)/2008+1+(x+5)/2007+1+(x+6)/2006+1)=0
(x+2012)/2011+(x+2012)/2010+(x+2012/2009-(x+2012)/2008-(x+2012)/2007-(x+2012)/2006=0
(x+2012)(1/2011+1/2010+1/2009-1/2008-1/2007-1/2006)=0
x+2012=0
x=-2012
KT
25 tháng 11 2017
1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009
DV
1
HT
1 tháng 9 2015
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}+\frac{x+3}{2009}\)
=> \(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}-\frac{x+3}{2009}=0\)
=> \(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
=> x + 3 = 0
=> x = 0 - 3
=> x = -3
U
28 tháng 8 2018
\(a,\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
\(\Rightarrow\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)+\left(\frac{x+7}{2008}+1\right)=0\)
\(\Rightarrow\frac{x+2016}{2010}+\frac{x+2016}{2009}+\frac{x+2006}{2008}=0\)
chỉ bt lm v thoi "(
KT
28 tháng 8 2018
a) \(\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
<=> \(\frac{x+5}{2010}+1+\frac{x+6}{2009}+1+\frac{x+7}{2008}+1=0\)
<=> \(\frac{x+2015}{2010}+\frac{x+2015}{2009}+\frac{x+2015}{2008}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
<=> \(x+2015=0\) (do 1/2010 + 1/2009 + 1/2008 # 0 )
<=> \(x=-2015\)
Vậy...
b) mạo phép chỉnh đề
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+344}{5}=0\)
<=> \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+344}{5}-3=0\)
<=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{5}=0\)
làm tương tự a
NT
8
25 tháng 9 2016
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)\)
\(\Rightarrow\left(\frac{x+1}{2009}+\frac{2009}{2009}\right)+\left(\frac{x+2}{2008}+\frac{2008}{2008}\right)=\left(\frac{x+3}{2007}+\frac{2007}{2007}\right)+\left(\frac{x+4}{2006}\frac{2006}{2006}\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2007}+\frac{x+2010}{2006}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2007}-\frac{x+2010}{2006}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
=>x+2010=0
=>x=-2010
Vậy x = -2010
LT
1
17 tháng 9 2017
Trừ 1 đi ở mỗi phân số, ta có:
\(\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Rightarrow\frac{x-1}{2009}-\frac{2009}{2009}+\frac{x-2}{2008}-\frac{2008}{2008}=\frac{x-3}{2007}-\frac{2007}{2007}+\frac{x-4}{2006}-\frac{2006}{2006}\)
\(\Rightarrow\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}=\frac{x-3-2007}{2007}+\frac{x-4-2006}{2006}\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Rightarrow\left[x-2010\right]\left[\frac{1}{2009}+\frac{1}{2008}\right]=\left[x-2010\right]\left[\frac{1}{2007}+\frac{1}{2006}\right]\)
Sẽ có hai trường hợp
TH1: Cả hai vế đều bằng 0
Ta có: \(\hept{\begin{cases}\frac{1}{2009}+\frac{1}{2008}\ne0\\\frac{1}{2007}+\frac{1}{2006}\ne0\end{cases}}\Rightarrow x-2010=0\Rightarrow x=2010\)
TH2: Cả hai vế khác 0
Ta bỏ đi x - 2010 vì cả hai bên đều có
\(\Rightarrow\frac{1}{2009}+\frac{1}{2008}=\frac{1}{2007}+\frac{1}{2006}\)Vô lí
Vậy x = 2010