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

bố tớ làm giáo viên, bảo bài này đúng đó

\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}=\dfrac{x-3}{2012}+\dfrac{x-4}{2011}\)

\(\Leftrightarrow\text{​​}\text{​​}\text{​​}\dfrac{x-1}{2014}-1+\dfrac{x-2}{2013}-1=\dfrac{x-3}{2012}-1+\dfrac{x-4}{2011}-1\)

\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}-\dfrac{x-2015}{2012}-\dfrac{x-2015}{2011}=0\)

\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)

mà \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\)

nên \(x-2015=0\)

\(\Leftrightarrow x=2015\)

bấm vào chữ xanh này nha bn : /hoi-dap/question/113985.html

(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

đúng ko các bạn 

ủng hộ mình lên 280 điểm với các bạn

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\) 

=> \(x+2014=0\) 

                  \(x=0-2014\) 

                  \(x=-2014\)

\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)

\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)

\(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\frac{x-3}{2011}+1-\frac{x-4}{2010}+1=0\)

\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-3}{2011}-1\right)-\left(\frac{x-4}{2010}-1\right)=0\)

\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)

\(\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

\(x-2014=0:\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)\)

\(x-2014=0\)

\(x=2014\)

Nhớ tk cho mình nha =3

A.R.M.Y   FIGHTING!!!!

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`

`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`

`=>x=-2014`

Ta có: \(\frac{x-1}{2013}+\frac{x-2}{2012}=\frac{x-3}{2011}+\frac{x-4}{2010}\)

\(\Rightarrow\frac{x-1}{2013}+1+\frac{x-2}{2012}+1=\frac{x-3}{2011}+1+\frac{x-4}{2010}+1\)

\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}=0\)

\(\Rightarrow\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\ne0\)

=> x - 2014 =0

=> x          = 2014

Vậy x = 2014