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Ta có: \(2-x+2005=1-x+2006=-x+2007\)
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}+1-2=\frac{1-x}{2006}+1+\left(\frac{-x}{2007}+1\right)-2\)
\(\Leftrightarrow\frac{2007-x}{2005}=\frac{2007-x}{2006}+\frac{2007-x}{2007}\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Rightarrow2007-x=0\)
\(\Rightarrow x=2007\)
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}-\frac{1-x}{2006}+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\frac{2-x}{2005}+1-\frac{1-x}{2006}-1+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\left(\frac{2-x}{2005}+1\right)-\left(\frac{1-x}{2006}+1\right)-\left(1-\frac{x}{2007}\right)=0\)
\(\Leftrightarrow\frac{2-x+2005}{2005}-\frac{1-x+2006}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\frac{2007-x}{2005}-\frac{2007-x}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Leftrightarrow2007-x=0\) < Vì \(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\ne0\)>
\(\Leftrightarrow x=2007\)
VẬY \(x=2007\)
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)
\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)(2)
Từ (1) và (2) \(\Rightarrow x+2009=0\)\(\Rightarrow x=-2009\)
Vậy \(x=-2009\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
3 hạng tử đầu , mỗi hạng tử cùng cộng 1
Hạng tử cuối trừ 3
Nhân tử chung : x + 2010
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
theo đề baiif nên
x+1/2008+x+2/2007+x+3/2006-(x+4/2005)-(x+5/2004)-(x+6/2003)=0
suy ra [(x+1/2008)+1]+[(x+2/2007)+1]+[x+3/2006)+1]-[(x+4/2005)+1]-[(x+5/2004)+1]-[(x+6/2003)+1]=0
suy ra (x+2009/2008)+(x+2009/2007)+(x+2009/2006)-(x+2009/2005)-(x+2009/2004)-(x+2009/2003)=0
nên (x+2009)(1/2008+1/2007+1/2006-1/2005-1/2004-1/2003)=0
V1 V2
Dễ thấy V2>0 NÊN x+2009=0 suy ra x=-2009
a) \(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
\(\frac{8-6x-4+x}{10}=\frac{5x+10}{10}\)
\(4-5x=5x+10\)
\(4-5x-5x-10=0\)
\(-6-10x=0\)
\(\Rightarrow x=\frac{-3}{5}\)
Vậy....
\(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
\(\Leftrightarrow\)\(\frac{2.\left(4-3x\right)}{10}-\frac{4-x}{10}=\frac{5.\left(x+2\right)}{10}\)
\(\Rightarrow\) 2.( 4 - 3x ) - 4 + x = 5.( x + 2 )
\(\Leftrightarrow\)8 - 6x - 4+ x = 5x + `10
\(\Leftrightarrow\)-6x + x - 5x = -8 + 4 + 10
\(\Leftrightarrow\) -10x = 6
\(\Leftrightarrow\)\(x=\frac{-3}{5}\)
Vậy phương trình có nghiệm là: \(x=\frac{-3}{5}\)
b ) \(\frac{x+1}{2009}+\frac{x+2}{2008}=\frac{x+2007}{3}+\frac{x+2006}{4}\)
\(\Leftrightarrow\) \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1\)\(=\frac{x+2007}{3}+1+\frac{x+2006}{4}+1\)
\(\Leftrightarrow\)\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}\)\(=\frac{x+2007}{3}+\frac{3}{3}+\frac{x+2006}{4}+\frac{4}{4}\)
\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{3}+\frac{x+2006}{4}\)
\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{3}-\frac{x+2010}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+2010=0\) ( Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\ne0\))
\(\Leftrightarrow\) \(x=-2010\)
Vậy phương trình có nghiệm là: x = -2010
Có : \(\frac{2-x}{2005}-1\) = \(\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\) \(\frac{2-x}{2005}+1=\left(\frac{1-x}{2006}+1\right)-\left(\frac{x}{2007}+1\right)\)
\(\Leftrightarrow\) \(\frac{-x+2007}{2005}=\frac{-x+2007}{2006}+\frac{-x+2007}{2007}\)
\(\Leftrightarrow\) \(\left(-x+2007\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Leftrightarrow\) \(-x+2007=0\) ( vì \(\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\ne0\)
\(\Leftrightarrow\) \(x=2007\)
Vậy \(x=2007\)
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