X=2012

X * 2012 - X=2012*2010+2012

=>x*(2012-1)=2012*(2010+1)

=>x*2011=2012*2011

=>x=2012 (2 vế chia 2011)

2012

X x 2012 - X = 2012 x 2010 + 2012

X x ( 2012 - 1 ) = 2012 x ( 2010 + 1 )

X x 2011 = 2012 x 2011

\(\Rightarrow\)X = 2012

 X x 2012 - X = 2012 x 2010 + 2012

<=>2011xX=2012x2011

<=>X=2012

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

 ~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~

y x 2012 - y = 2012 x 2010 + 2012 

2011 x y = 4046132

           y = 4046132 : 2011

           y = 2012

vậy y = 2012

cbht

Ta có:

y x 2012 - y = 2012 x 2010 + 2012

y*2012 - y*1=2012*2010+2012*1

y*(2012- 1)=2012*(2010+1)

y*2011=2012*2011

y=2012(bớt mỗi vế 1 thừa số 2011)

Vậy y=2012

TL:

y x 2012 - y = 2012 x 2010 + 2012

<=> 2011y = 2012 x 2011

<=> y = 2012 x 2011 : 2011

<=> y = 2012

#study well#

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

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)

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 \(x=2013\)

Ta có : 

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

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

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

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

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

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

Nên \(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

Vậy \(x=2013\)

Chúc bạn học tốt ~ 

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

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

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

\(\Leftrightarrow x=2013\)