ta có : x=2010

->x-1=2009

A(x)=x²⁰¹⁰-(x-1).x²⁰⁰⁹ -(x-1).x²⁰⁰⁸ -...-(x-1).x+1

A(x)=x²⁰¹⁰-x²⁰¹⁰+x²⁰⁰⁹-x²⁰⁰⁹+x²⁰⁰⁸-...-x²+x+1

A(x)=x+1=2010+1=2011

Cảm ơn

     P(0)=2

      => a.0²+b.0+c= -2

      = >c= -2

4P(x)=4.(ax²+bx-2)                      P(2x-1)=a(2x-1)²+b(2x-1)-2

        =4ax²+4bx-8                                  =a(4x²  -4x+1)+b(2x-1)-2

4P(x)-P(2x-1)=6x-6

=>(4ax²+4bx-8)-[a(4x²-4x+1)+b(2x-1)-2]=6x-6

=>4ax²+4bx+8-(4ax²-4ax+a+2bx-b-2)=6x+6

=>4ax²+4bx+8-4ax²+4ax-a-2bx+b+2=6x+6

=>(4ax²-4ax²)+(4bx-2bx)+(8+2)+4ax-a-b=6x+6

=>(2bx-b)+(4ax-a)+10=6x+6

=>b(2x-1)+(2ax-a)+2ax=6x+(6-10)

=>b(2x-1)+a(2x-1)+2ax=6x-4

=>(a+b).(2x-1)+2ax=3(2x-1)-1

=>3(2x-1)-(a+b).(2x-1)=1-2ax

=>(3-a-b).(2x-1)=1-2ax

=>(3-a-b).(2x-1)=a(2x-1)+a-1

=>(3-a-a-b).(2x-1)=a-1

=>(3-2a-b).(2x-1)=a-1......

@vũ thị thu thao

Bạn giải tiếp đi!!!! Mình thấy bạn làm đúng nên k rồi đó

pan a ban giong bup be lam nhung bup be lam = nhua deo va no del co nao nhe

A(2010)=x^2010 - 2009x^2009 - 2009x^2008 - 2009x^2007 -...- 2009x + 1

ta có: 2010-1=2009 --> x-1=2009

thay x-1=2009 vào đa thức A(2010) ta được:

A(2010)=x^2010 - x^2009(x-1) - x^2008(x-1) - x^2007(x-1) -...- x(x-1) + 1

=x^2010 - x^2010 + x^2009 - x^2009 + x^2008 - x^2008 + x^2007 -...- x^2 + x + 1 

= x + 1 

thay x=2010 vao x+1 ta được:

2010+1=2011

vậy A(2010)=2011

Ta có:

\(P\left(0\right)=a.0^2+b.0+c=-2\Rightarrow c=-2\)

\(4P\left(x\right)-P\left(2x-1\right)=6x-6\)

\(\Leftrightarrow4\left(ax^2+bx+c\right)-a\left(2x-1\right)^2-b\left(2x-1\right)-c=6x-6\)

\(\Leftrightarrow4ax^2+4bx+4c-a\left(4x^2-4x+1\right)-2bx+b-c=6x-6\)

\(\Leftrightarrow2bx-8+4ax-a+b+2=6x-6\)(Do c=-2)

\(\Leftrightarrow x\left(2b+4a\right)-\left(a-b+6\right)=6x-6\)

Do thỏa mãn với mọi x nên:

\(\left\{{}\begin{matrix}2b+4a=6\\a-b+6=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b+2a=3\\a-b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

Ta có :

\(\left\{{}\begin{matrix}a=1\\b=1\\c=-2\end{matrix}\right.\Leftrightarrow a+b+c=0\)

=> \(P\left(x\right)=x^2+x-2\)