Tìm x biết: x+1/2013 + x+2/2012 + x+3/2011 = -3
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TK
0
CM
0
TL
1
NX
0
TN
3
7 tháng 9 2015
(x + 4)/2010 + (x+3)/2011 = (x+2)/2012 + (x+1)/2013
<=> [(x + 4)/2010 + 1] + [(x+3)/2011 + 1] = [(x+2)/2012 + 1] + [(x+1)/2013 + 1]
<=> (x + 2014)/2010 + (x + 2014)/2011 = (x + 2014)/2012 + (x + 2014)/2013
<=> (x + 2014)/2010 + (x + 2014)/2011 - (x + 2014)/2012 - (x + 2014)/2013 = 0
<=> (x + 2014).(1/2010 + 1/2011 - 1/2012 - 1/2013) = 0
Ta thấy (1/2010 + 1/2011 - 1/2012 - 1/2013) ≠ 0
Vậy suy ra x = -2014
IL
0
HT
1
9 tháng 8 2018
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)
\(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\frac{x-3}{2011}+1-\frac{x-4}{2010}+1=0\)
\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-3}{2011}-1\right)-\left(\frac{x-4}{2010}-1\right)=0\)
\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)
\(\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(x-2014=0:\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)\)
\(x-2014=0\)
\(x=2014\)
Nhớ tk cho mình nha =3
A.R.M.Y FIGHTING!!!!
\(\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}=-3\)
\(=>\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}+3=0\)
\(=>\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=0\)
\(=>\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=0\)
\(=>\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
=>x+2014=0 (vì \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\) khác 0)
=>x=-2014