`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`

`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`

`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`

`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)

`<=>x=2014`

Vậy `S={2014}`.

=>x-2014=0

hay x=2014

\(\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2019}-1+\dfrac{x-5}{2010}-1+\dfrac{x-6}{2008}-1\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\Leftrightarrow x=2014\)

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

\(\Leftrightarrow\left(x-2014\right).A=0\)

\(\text{Vì A }\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)

<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)

<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)

do 1/2012+1/2011....+1 khác 0 =>x-2013=0<=>x=2013

vậy..........................

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

\(\left(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}\right)-2012=0\)

\(\Rightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)

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

Vì \(x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)nên x - 2013 hoặc \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\) = 0. Nhưng \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\ne0\) nên x - 2013 = 0. Vì vậy x = 2013.

Vậy...

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

\(\Leftrightarrow\left(\dfrac{x-1}{2012}-1\right)+\left(\dfrac{x-2}{2011}-1\right)+\left(\dfrac{x-3}{2010}-1\right)+...+\left(\dfrac{x-2012}{1}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)

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

Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...1\ne0\)

\(\Rightarrow x-2013=0\)

\(\Leftrightarrow x=2013\)

Vậy...

Lời giải:

Ta có:

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

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

\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

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

Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)

\(\Leftrightarrow x=2013\)

Vậy PT có nghiệm \(x=2013\)

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)

\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)

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

Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)

\(\Rightarrow x-2013=0\Rightarrow x=2013\)

đề sai?

\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)

\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x+2013=0\)

\(\Leftrightarrow x=-2013\)

Bài của bạn nè bạn gái!

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{1012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{10}{2008}\ne0\)

\(\Rightarrow x-2014=0\Rightarrow x=2014\)

vậy x=2014

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\dfrac{x-1}{2013}+1+\dfrac{x-2}{2012}+1+\dfrac{x-3}{2011}+1-\dfrac{x-4}{2010}+1-\dfrac{x-5}{2009}+1-\dfrac{x-6}{2008}+1=0\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Vậy PT có nghiệm là \(x=2014\)

Trừ 1 vào mỗi phân thức ở vế trái, tương tự với vế phải sẽ xuất hiện tử thức chung x-2014 rồi đặt nó làm nhân tử chung => x=2014