2012 . | x - 2011| + (x-2011)² = 2013 . | 2011 - x|

|x-2011|.|x-2011| + 2012 . | x - 2011| - 2013 . | 2011- x| =0

|x - 2011|.| x - 2011| + 2012 .| x - 2011| - 2013 | x - 2011| = 0

| x- 2011| .| x -2011|  - | x - 2011| = 0

| x - 2011|. { | x - 2011| - 1} = 0

\(\left[{}\begin{matrix}\left|x-2011\right|=0\\\left|x-2011\right|-1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2011\\x=2012\\x=2010\end{matrix}\right.\)

Kết luận x \(\in\) { 2010; 2011; 2012}

`x xx 2/3 xx 3/4 xx 4/5 xx ... xx 2010/2011 = 2/2012`

`<=> x/2011 = 1/1006`

`=> x = 2011/1006`

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

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

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

\(\Rightarrow x=-2014\)

\(2012.\left|x-2011\right|+\left(x-2011\right)^2=2013\left|2011-x\right|\)

\(2012.\left|x-2011\right|+\left|x-2011\right|^2=2013\left|x-2011\right|\)

\(\left|x-2011\right|\left(2012+\left|x-2011\right|\right)=2013\left|x-2011\right|\)

\(\Rightarrow2012+\left|x-2011\right|=2013\)

\(\left|x-2011\right|=1\)

\(\Rightarrow\orbr{\begin{cases}x=2012\\x=-2010\end{cases}}\)

dư 0

vì ta thấy:

2010+2013=2015 chia hết cho 5

nên A+B chia 5 dư 0

tíc mình nha

Ta có :

2012 x 2012 x 2012 x ... x 2012 x 2012 ( 2013 chữ số 2012 ) = 2012 ^ 2013 => chữ số tận cùng là 8                               (1)

2013 x 2013 x 2013 x ... x 2013 x 2013 ( 2012 chữ số 2013 ) = 2013 ^ 2012 => chữ số tận cùng là 9                               (2)

Từ (1) và (2) => ( 8 + 9 ) : 5 = 17 : 5 dư 2

Mơ đi nhé. Dư 3

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025+2025\cdot2012-2025\cdot2013}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025\left(1+2012-2013\right)}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025\cdot0}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{0}\)

\(A=0\)

=))

để mk sửa lại đề cho

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

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

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

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

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

\(=x^{2012}.0-x^{2011}.0+...+x.0+2012-1\)

=2011

Vậy f(2012)=2011