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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
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 \(x\in\left\{\varnothing\right\}\)
16 tháng 4 2019
a) \(\frac{5x-3}{5}+\frac{2x+1}{4}\le\frac{2-3x}{2}-5\)
\(\Leftrightarrow\frac{4\cdot\left(5x-3\right)}{20}+\frac{5\left(2x+1\right)}{20}\le\frac{10\left(2-3x\right)}{20}-\frac{20\cdot5}{20}\)
\(\Leftrightarrow20x-12+10x+5\le20-30x-100\)
\(\Leftrightarrow20x+10x+30x\le20-100+12-5\)
\(\Leftrightarrow60x\le-73\)
\(\Leftrightarrow x\le\frac{-73}{60}\)
7 tháng 4 2019
ta có:
\(\frac{x+2}{2013}+\frac{x+5}{2010}>\frac{x+8}{2007}+\frac{x+11}{2004}\)
\(\Leftrightarrow\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+5}{2010}+1\right)>\left(\frac{x+8}{2007}+1\right)+\left(\frac{x+11}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}>\frac{x+2015}{2007}+\frac{x+2015}{2004}\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}-\frac{x+2015}{2007}-\frac{x+2015}{2004}>0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}\right)>0\)
\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)
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
HT
1
30 tháng 8 2015
\(\left(x^{2007}+y^{2007}\right)\left(x^3+y^3\right)=1.\left(x^3+y^3\right)\)
=> \(x^{2010}+x^3y^{2010}+x^{2010}.y^3+y^{2010}=x^3+y^3\)
=> \(x^3y^3\left(x^{2007}+y^{2007}\right)=x^3+y^3-\left(x^{2010}+y^{2010}\right)\)
Vì \(x^3+y^3=x^{2010}+y^{2010}\Rightarrow x^3+y^3-\left(x^{2010}+y^{2010}\right)=0\)
<=> \(x^3y^3\left(x^{2007}+y^{2007}\right)=0\)
=> x^3 = 0 hoặc y^3 = 0 hoặc x^2007 + y^2007 = 0
(+) với x^3 = 0 => x = 0 => 0^2007 + y^2007 = 1 => y = 1
(+) với y^3 = 0 => x = 1
(+) x^2007 + y^2007 = 0 ( loại tái với đề bài x^2007 + y^2007 = 1 )