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\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}\right)=\left(x-23\right)\left(\frac{1}{26}+\frac{1}{27}\right)\text{ nhận thấy:}\frac{1}{24}+\frac{1}{25}>\frac{1}{26}+\frac{1}{27}\)
\(\Rightarrow x-23=0\Leftrightarrow x=23\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\Rightarrow\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)=\left(\frac{x+3}{2002}+1\right)+\left(\frac{x+4}{2001}+1\right)\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\text{dạng giống câu a rồi nha}\)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\Leftrightarrow300-x=0\)
Vậy: x=300
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
b. \(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1< \frac{x-2}{98}-1+\frac{x-4}{96}-1+\frac{x-6}{94}-1\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}-\frac{x-100}{98}-\frac{x-100}{96}-\frac{x-100}{94}< 0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Vì \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Nên \(x-100< 0\Leftrightarrow x< 100\)
P/S: Bài a có sai đề ko bn
Mình nhầm
Vì \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)< 0\)
Nên \(\left(x-100\right)>0\Leftrightarrow x>100\)
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
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
Bài làm
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x+2+2005}{2005}\right)+\left(\frac{x+3+2004}{2004}\right)+\left(\frac{x+4+2003}{2003}\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right).\frac{1}{2005}+\left(x+2007\right).\frac{1}{2004}+\left(x+2007\right).\frac{1}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2007=\frac{0}{\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}}\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy phương trình trên có tập nghiệm S = { -2007 }
# Học tốt #
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}>0\)(2)
Từ (1), (2) \(\Rightarrow x+2017=0\)\(\Leftrightarrow x=-2017\)
Vậy \(x=-2017\)
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\)
=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)
\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(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 \(x=-2009\)