Ta có: x=2011 \(\Rightarrow\)x+1=2012

\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)

=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)

= \(x-1=2011-1=2010\)

=

Thay 2012=x+1.

\(A=x^{2011}-\left(x+1\right)x^{2010}+\left(x+1\right)x^{2009}-\left(x+1\right)x^{2008}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(A=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^3-x^2+x^2+x-1\)

\(A=x-1=2011-1=2010\)

x=2012

nên x+1=2013

\(f\left(x\right)=x^{2013}-x^{2012}\left(x+1\right)+x^{2011}\left(x+1\right)-...+x\left(x+1\right)-1\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...+x^2+x+1\)

=x+1=2013

x=2012

nên x+1=2013

\(f\left(x\right)=x^{2013}-x^{2012}\left(x+1\right)+x^{2011}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-1\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}-...-x^3-x^2+x^2+x-1\)

=x-1

=2012-1=2011

a, Đ/k x-2012>=0 suy ra x>=2012

|x-2011|=\(\orbr{\begin{cases}x-2012\\2012-x\end{cases}}\)

TH1:x-2011=x-2012

suy ra 0=4023(loại vì mất x)

TH2: x-2011=2012-x

suy ra 2x=4023

suy ra x=2011,5

Vậy..........

k o  b i ế t  n h a

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

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

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

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

\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)

\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)

Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức

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

\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)

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

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

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

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

Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)

Nên \(x-2010=0\)

\(\Rightarrow\)\(x=2010\)

Vậy \(x=2010\)

Chúc bạn học tốt ~ 

\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)

\(\Rightarrow\left(\dfrac{x+4}{2009}+1\right)+\left(\dfrac{x+3}{2010}+1\right)=\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+1}{2012}+1\right)\)

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

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

Vì \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)

=> x +2013 = 0

=> x = -2013

\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)

\(\Leftrightarrow\dfrac{x+4}{2009}+1+\dfrac{x+3}{2010}+1=\dfrac{x+2}{2011}+1+\dfrac{x+1}{2012}+1\)

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

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

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

\(\Leftrightarrow x+2013=0\).Do \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)

\(\Rightarrow x+2013=0\)

\(\Leftrightarrow x=-2013\)