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KV
1
KH
18 tháng 2 2020
\(pt\Leftrightarrow\frac{x}{2009}+\frac{1}{2009}+\frac{x}{2008}+\frac{2}{2008}=\frac{x}{3}+\frac{2007}{3}+\frac{x}{4}+\frac{2006}{4}\Leftrightarrow\frac{x}{2009}+\frac{x}{2008}-\frac{x}{3}-\frac{x}{4}=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x=\frac{\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}}{\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}}=-2010\)
K
1
4 tháng 5 2019
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)
<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)
<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0
<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0
<=> x+2011=0
<=> x=-2011
Vậy pt có nghiệm là x=-2011
TD
1
22 tháng 2 2015
lấy mỗi phân số trừ đi 1 ta đc (x-1)/2009 -1 + (x-2)/2008 -1 = (x-3)/2007-1 + (x-4)/2006 -1
suy ra (x-2010)/2009 + (x-2010)/2008 - (x-2010)/2007 - (x-2010)/2006 = 0
đặt (x-2010) làm nhân tử chung ta được
(x-2010).(1/2009 + 1/2008 - 1/2007 - 1/2006 ) = 0
=> x-2010 = 0
=> x = 2010
PN
11 tháng 2 2016
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\) \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\) \(\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Leftrightarrow\) \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)'
Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\) nên \(x-2010=0\) \(\Leftrightarrow\) \(x=2010\)
Vậy, tập nghiệm của pt là \(S=\left\{2010\right\}\)
10 tháng 7 2019
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....
NN
10 tháng 7 2019
\(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
\(\Leftrightarrow\)\(\frac{2.\left(4-3x\right)}{10}-\frac{4-x}{10}=\frac{5.\left(x+2\right)}{10}\)
\(\Rightarrow\) 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
DM
13 tháng 2 2018
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
7 tháng 1 2016
(x+1)/2011+1+(x+2)/2010+1+(x+3)/2009+1-((x+4)/2008+1+(x+5)/2007+1+(x+6)/2006+1)=0
(x+2012)/2011+(x+2012)/2010+(x+2012/2009-(x+2012)/2008-(x+2012)/2007-(x+2012)/2006=0
(x+2012)(1/2011+1/2010+1/2009-1/2008-1/2007-1/2006)=0
x+2012=0
x=-2012
7 tháng 3 2018
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=1+1+1-3\)
\(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+1\right)=0\)
Mà \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
Vậy ..
\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\right)=0\Leftrightarrow x=2010\)
\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
=>x-2010=0
hay x=2010