a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=2012\\\left(x-2012-x+2011\right)\left(x-2012+x-2011\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2012\\2x=2023\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

b: Trường hợp 1: x<2010

Pt sẽ là 2010-x+2011-x=2012

=>4021-2x=2012

=>2x=2009

hay x=2009/2(nhận)

TRường hợp 2: 2010<=x<2011

=>x-2010+2011-x=2012

=>1=2012(vô lý)

Trường hợp 3: x>=2011

=>x-2010+x-2011=2012

=>2x=2012+4021=6033

hay x=6033/2(nhận)

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.

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

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

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

mình giải ra trc mà, k mik chứ

suy ra hai truong hop

1 : x-2011=x-2012

suy ra x-x=2011-2012(loai)

2 : x-2011=-(x-2012)

suy ra : x-2011=-x+2012                                

2x=2011+2012

2x=4023

x=2011.5

kho..................lam............................tich,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,minh..........................troi........................ret............................wa.................ung ho minh.................hu....................hu..............hu................hat..............hat....................s

Cau a biet lam roi de lam cau b cho:

Ta thay:tong 2 so hang do la tong 2 so tu nhien lien tiep nen se la so le (*) ma 2012 la so chan(trai voi "*")

=>Khong ton tai so nguyen x thoa man

Vay khong ton tai so nguyen x thoa man

a/ \(\left|x-2011\right|=x-2012\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2011=x-2012\\x-2011=-x+2012\end{matrix}\right.\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x-x=-2012+2011\\x+x=2012+2011\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=-1\left(loại\right)\\2x=4023\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{4023}{2}\)

Vậy ...

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 \(x-2011=0\)

\(x=2011\)