|2013x² + 4026x| = |x + 2|

<=> |2013x(x + 2)| - |x + 2| = 0

<=> |2013x|.|x + 2| - |x + 2| = 0

<=> |x + 2|(|2013x| - 1) = 0

<=> |x + 2|=0 hoặc |2013x|-1=0

<=> x+2=0 hoặc 2013x=1 hoặc 2013x=-1

=> x=-2 hoặc x=1/2013 hoặc x=-1/2013

Vậy ...

\(\Leftrightarrow2013\left(x^2-2x+1\right)+2014\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow2013\left(x-1\right)^2+2014\left(y+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Lời giải:

Tại $x=2012$ thì $x-2012=0$. Ta có

$P(x)=x^5-2013x^4+2013x^3-2013x^2+2013x-2014$
$=x^4(x-2012)-x^3(x-2012)+x^2(x-2012)-x(x-2012)+(x-2012)-2$

$=(x-2012)(x^4-x^3+x^2-x+1)-2$

$=0.(x^4-x^3+x^2-x+1)-2=-2$

Cách khác:

Ta có: x=2012

nên x+1=2013

Ta có: \(P\left(x\right)=x^5-2013x^4+2013x^3-2013x^2+2013x-2014\)

\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2014\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2014\)

\(=x-2014=2012-2014=-2\)

Đề hơi sai sai bạn ơi

2013x2+98x2013=2013X{2+98}=2013X100=201300

=2013x(2+98)

=2013x100

=201300

Tk plz 

thanks

2013 x 2 + 98 x 2013 

= 2013 x ( 98 + 2 )

= 2013 x 100

= 201300

ai k mình mình k lại nha 

=>2013|x||x+2|-|x+2|=0

=>|x+2|(2013|x|-1)=0

=>x+2=0 hoặc 2013|x|=1

hay \(x\in\left\{-2;\dfrac{1}{2013};-\dfrac{1}{2013}\right\}\)