Tìm x, biết: (x+4)/2012+(x+3)/2013=(x+2)/2014+(x+1)/2015
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DV
0
NL
2
IM
10 tháng 9 2016
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)
\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
LM
1
TB
12 tháng 4 2017
CÓ: \(\frac{x-1}{2015}+\frac{x-2}{2014}-\frac{x-3}{2013}-\frac{x-4}{2012}=0\)\(0\)
<=>\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)=0\)
<=>\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)
<=>\(\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Do:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
=>\(x-2016=0\)
<=>\(x=2016\)
DM
1
TT
4
10 tháng 4 2016
\(\frac{x+3}{2013}+1+\)\(\frac{x+4}{2012}+1+\frac{x+5}{2011}+1\)=\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x}{2016}+1\)
\(\Rightarrow\frac{x+2016}{2013}+\frac{x+2016}{2012}+\frac{x+2016}{2011}=\frac{x+2016}{2014}+\frac{x+2016}{2016}\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2016}=0\right)\)
Vì 1/2016+...+1/2011>0 nên (x+2016)=0
suy ra x= -2016
nếu đúng xin kết bạn
ch
TT
0
ZZ
0
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
=> \(\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)
=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}>\frac{1}{2014};\frac{1}{2013}>\frac{1}{2015}\)
=> \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
=> \(x+2016=0\)
=> \(x=-2016\)