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6 tháng 7 2019
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....
KC
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HH
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1 tháng 3 2019
https://olm.vn/hoi-dap/detail/212443421285.html
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TT
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4 tháng 2 2017
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\
\)
Cộng từng hạng tử của hai vế với 1
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}=\frac{x+3+2002}{2002}+\frac{x+4+2001}{2001}\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)\ne0\)nên \(x+2005=0\Rightarrow x=-2005\)
Phương trình có nghiệm duy nhất: x=2005
4 tháng 2 2017
(x+1)/2004+(x+2)/2003=(x+3)/2002+(x+4)/2001
(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)/2002=0
(x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=>x+2005=0
=>x=-2005
23 tháng 2 2020
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 .
4 tháng 4 2016
2 -x/2002 + 1 -1 = 1-x/2003 + 1 - x/2004 + 1
=> 2004 - x/ 2002 = 2004 - x/ 2003 + 2004 -x/2004
=> (2004 -x) ( 1/2002-1/2003-1/2004)
ta thấy ( 1/2002-1/2003-1/2004) # 0
=> 2004 -x = 0 => x = 2004
17 tháng 2 2020
a)\(\frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\Rightarrow300-x=0\Rightarrow x=300\)
b)\(\frac{2-x}{2002}+1=\frac{1-x}{2003}+2-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{1-x}{2003}+1+1-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}+\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Mà \(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)
c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}-2=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-2\)
\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)
Mà\(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\ne0\)
\(\Rightarrow x^2-10x-2000=0\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)