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T
17 tháng 9 2018
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
9 tháng 10 2016
\(\frac{x+4}{2011}+\frac{x+3}{2012}=\frac{x+2}{2013}+\frac{x+1}{2014}\)
\(\Leftrightarrow\left(\frac{x+4}{2011}+1\right)+\left(\frac{x+3}{2012}+1\right)-\left(\frac{x+2}{2013}+1\right)-\left(\frac{x+1}{2014}+1\right)=0\)
\(\Leftrightarrow\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow x+2015=0\) (Vì: \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\) )
\(\Leftrightarrow x=-2015\)
NK
0
AD
1
NT
1
23 tháng 2 2017
trước tiên bạn phải tính:
2013/1+2012/2+2011/3+.....+2/2012+1/2013
=1+2012/2)+(1+2011/3)+.....+(1+2/2012)+(1+1/2013) +1 {BƯỚC NÀY TÁCH 2013 RA LÀM 2013SỐ1 ĐỂ CÔNG VS CÁC THỪA SỐ CÒN LẠI}
=2014/2+2014/3+...+2014/2012+2014/2013+2014/2014
=2014.(1/2+1/3+....+1/2012+1/20131/2014
suy ra x=2014
7 tháng 1 2019
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}=\dfrac{x-3}{2012}+\dfrac{x-4}{2011}\)
\(\Leftrightarrow\text{}\text{}\text{}\dfrac{x-1}{2014}-1+\dfrac{x-2}{2013}-1=\dfrac{x-3}{2012}-1+\dfrac{x-4}{2011}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}-\dfrac{x-2015}{2012}-\dfrac{x-2015}{2011}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
mà \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\)
nên \(x-2015=0\)
\(\Leftrightarrow x=2015\)
20 tháng 3 2019
\(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)
\(\Leftrightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)
\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}=\frac{x+2014}{2012}+\frac{x+2014}{2013}\)
\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}-\frac{x+2014}{2012}-\frac{x+2014}{2013}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
V...
NT
3
I
20 tháng 10 2017
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)
\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)
\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)
\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)
\(\Rightarrow x-2015=0\)
\(x=0+2015\)
\(x=2015\)
=> \(\left(\frac{x+4}{2011}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+1}{2014}+1\right)\)
=> \(\frac{x+5}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)
=> \(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
=> x = -2015 Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)