Giải phương trình:
\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}\)
K
Khách
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2 tháng 4 2020
\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))
\(\Leftrightarrow x=-2004\)
2 tháng 4 2020
\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)
\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)
\(\Leftrightarrow x=-2004\)
Vậy PT có tập nghiệm S = {-2004}
FB
1
3 tháng 2 2020
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2001}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\left(2003-x\right)=0\) (vì \(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))
\(\Leftrightarrow x=2003\).
Vậy tập nghiệm của phương trình là \(S=\left\{2003\right\}\).
NT
0
10 tháng 5 2018
a,\(\Leftrightarrow\left(\frac{1-x}{2013}+1\right)=\left(\frac{2-x}{2012}+1\right)-\left(1-\frac{x}{2014}\right)\)
\(\Leftrightarrow\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{2014-x}{2014}\)
\(\Leftrightarrow\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0
\(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2014\left(do.cái.còn.lại.\ne0\right)\)
b,tương tự +1 vào cái thứ nhất ,+1 vào cái thứ 2,1- vào cái thứ 3 được x=2013
NN
13 tháng 11 2020
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)
\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(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 \(x=-2009\)
H
7 tháng 3 2018
Bạn chuyển về 1 vế sau đó trừ 1 vào mỗi phân thức ta được :
\(\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
Vì biểu thức bên phải khác 0 nên : \(x-2005=0\)=> \(x=2005\)
IA
23 tháng 3 2020
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
\(\Leftrightarrow\frac{x-2005}{2000}+\frac{x-2005}{2001}+\frac{x-2005}{2002}=\frac{x-2005}{2003}+\frac{x-2005}{2004}+\frac{x-2005}{2005}\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
<=> x - 2005 = 0
<=> x = 2005
Vậy ...............
2 tháng 3 2018
2.
pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0
<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0
<=> (x-2000).(1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004) = 0
<=> x-2000=0 ( vì 1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004 > 0 )
<=> x=2000
Tk mk nha
2 tháng 3 2018
1.
a, = (2x-1)^2-2.(2x-1)+1-4
= (2x-1-1)^2-4
= (2x-2)^2-4
= (2x-2-2).(2x-2+2)
= 2x.(2x-4)
b, = [x.(x+3)].[(x+1).(x+2)]
= (x^2+3x).(x^2+3x+1)-8
= (x^2+3x+1)^2-1-8
= (x^2+3x+1)^2-9
= (x^2+3x+1-3).(x^2+3x+1+3)
= (x^2+3x-2).(x^2+3x+4)
= ((x+1).(x+3).(x^2+3x-2)
Tk mk nha
\(pt\Leftrightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}.\)
\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}#0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy nghiệm của pt là x = -1