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\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
tự làm nốt~
kudo shinichi làm sai ở chỗ:
\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé
\(\frac{x+1}{2011}+\frac{x+2}{2010}=\frac{x+3}{2009}+\frac{x+4}{2008}\Leftrightarrow\frac{x+1}{2011}+1+\frac{x+2}{2010}+1=\frac{x+3}{2009}+1+\frac{x+4}{2008}+1\)
\(\Leftrightarrow\frac{x+1}{2011}+\frac{2011}{2011}+\frac{x+2}{2010}+\frac{2010}{2010}=\frac{x+3}{2009}+\frac{2009}{2009}+\frac{x+4}{2008}+\frac{2008}{2008}\)
\(\Leftrightarrow\frac{x+1+2011}{2011}+\frac{x+2+2010}{2010}=\frac{x+3+2009}{2009}+\frac{x+4+2008}{2008}\)
\(\Leftrightarrow\frac{x+2012}{2011}+\frac{x+2012}{2010}=\frac{x+2012}{2009}+\frac{x+2012}{2008}\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2012\right)\left(\frac{1}{2009}+\frac{1}{2008}\right)\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}=0\right)\)
mà 1/2011+1/2010-1/2009-1/2008 khác 0
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
\(\left(3x-2\right)^2-x\left(9x-2\right)=24\Leftrightarrow9x^2-12x+4-9x^2+2x=24\)
\(\Leftrightarrow-10x+4=24\Leftrightarrow-10x=20\Leftrightarrow x=-2\)
1; Ta có : x+1/2011 + x+2/2010 = x+3/2009 + x+4/ 2008
Suy ra: 2+(x+1/2011 + x+2/2010 ) = 2+( x+3/2009 + x+4/2008)
suy ra ban tach 2=1+1 roi cong 1 voi tưng phân số trên nha sẽ ra kết quả ngay thôi
2; gợi ý nè : (3x-2)^2 =(3x)^2 + 2*3x*2+2^2
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
Ta có: \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
=>\(\frac{2-x}{2007}=\frac{1-x}{2008}-\frac{x}{2009}+1\)
=>\(\frac{2-x}{2007}=\left(\frac{1-x}{2008}+1\right)-\frac{x}{2009}+1-1\)
=>\(\frac{2-x}{2007}+1=\frac{1-x+2008}{2008}+\left(1-\frac{x}{2009}\right)\)
=>\(\frac{2-x+2007}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)
=>\(\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)
=>\(\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)
=>\(\left(2009-x\right).\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)
=>2009-x=-
=>x=2009
Vậy tập nghiệm của phương trình S=2009
