Bài 1 : làm tương tự với bài 2;3 nhé

Ta có : \(f\left(0\right)=c=2010;f\left(1\right)=a+b+c=2011\)

\(\Rightarrow f\left(1\right)=a+b=1\)

\(f\left(-1\right)=a-b+c=2012\Rightarrow f\left(-1\right)=a-b=2\)

\(\Rightarrow a+b=1;a-b=2\Rightarrow2a=3\Leftrightarrow a=\dfrac{3}{2};b=\dfrac{3}{2}-2=-\dfrac{1}{2}\)

Vậy \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)

\(g\left(x\right)=ax^3-bx\)

\(f\left(x\right)=g\left(x\right)-15\)

\(f\left(-x\right)=-g\left(x\right)-15\)

\(f\left(x\right)+f\left(-x\right)=-30\)

\(f\left(5\right)+f\left(-5\right)=-30\Rightarrow f\left(-5\right)=-30-5=-35\)

\(f\left(-5\right)=-35\)

Bài 1 : 

\(P\left(0\right)=d=2017\)

\(P\left(1\right)=a+b+c+d=2\Rightarrow a+b+c=-2015\)(*)

\(P\left(-1\right)=-a+b-c+d=6\Rightarrow-a+b-c=6-2017=-2023\)(**)

\(P\left(2\right)=8a+4b+2c+d=-6033\Rightarrow8a+4b+2c=-8050\)

Lấy (*) + (**) ta được : \(2b=-4038\Rightarrow b=-2019\)

Thay vào (*) ta được \(a+c=4\)(***)

Lại có : \(8a+4b+2c=-8050\Rightarrow8a+2c=-8050+8076=26\)(****) 

(***) => \(8a+8c=32\)(*****)

Lấy (****) - (*****) => \(-6c=-6\Rightarrow c=1\Rightarrow a=3\)

Vậy  ....

MỌI NGƯỜI GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP LẮM Ạ.

\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)

\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)

\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)

\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)

\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)

\(\Rightarrow a-b=2\Rightarrow a=b+2\)

Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)

\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)

\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)

\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)

Ta có : \(f\left(0\right)=c=1\)

\(f\left(1\right)=a+b+c=2\)

\(f\left(2\right)=4a+2b+c=8\)

\(\Rightarrow c=1,a=\frac{5}{2},b=\frac{-3}{2}\)

Vì vậy mà \(f\left(x\right)=\frac{5}{2}x^2-\frac{3}{2}x+1\)

nên \(f\left(-2\right)=\frac{5}{2}.\left(-2\right)^2-\frac{3}{2}.\left(-2\right)+1=14\)

Ta có: \(f\left(0\right)=a.0^2+b.0+c=0+0+c=c\) mà \(f\left(0\right)=1\)\(\Rightarrow c=1\)

\(f\left(1\right)=a.1^2+b.1^2+c=a+b+1\)mà \(f\left(1\right)=2\)\(\Rightarrow a+b+1=2\)\(\Rightarrow a+b=1\)

\(f\left(2\right)=a.2^2+2.b+c=4a+2b+1\)mà \(f\left(2\right)=8\)\(\Rightarrow4a+2b+1=8\)\(\Rightarrow4a+2b=7\)\(\Rightarrow2\left(2a+b\right)=7\)\(\Rightarrow2a+b=3,5\)\(\Rightarrow a+\left(a+b\right)=3,5\)\(\Rightarrow a+1=3,5\)\(\Rightarrow a=2,5\)

Lại có: \(a+b=1\)\(\Rightarrow2,5+b=1\)\(\Rightarrow b=1-2,5=-1,5\)

Ta có: \(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=2,5.4+\left(-1.5\right).\left(-2\right)+1=10+3+1=14\)