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

\(P\left(0\right)=a\cdot0^3+b\cdot0^3+c\cdot0+d=2017\)

\(\Leftrightarrow d=2017\)

\(P\left(1\right)=a\cdot1^3+b\cdot1^2+c\cdot1+d=2\)

\(\Leftrightarrow a+b+a+d=2\)

\(P\left(-1\right)=a\cdot\left(-1\right)^3+b\cdot\left(-1\right)^2+c\cdot\left(-1\right)+d=6\)

\(\Leftrightarrow-a+b-c+d=6\)

\(P\left(2\right)=a\cdot2^3+b\cdot2^2+c\cdot2+d=-6033\)

\(\Leftrightarrow8a+4b+2c+d=-6033\)

Ta có

\(F\left(0\right)=2016\)

\(\Leftrightarrow a\cdot0^2+b\cdot0+c=2016\)

\(\Leftrightarrow0+0+c=2016\)

\(\Leftrightarrow c=2016\)

\(F\left(1\right)=2016\)

\(\Leftrightarrow a\cdot1^2+b\cdot1+c=2017\)

\(\Leftrightarrow a+b+c=2017\)

\(\Leftrightarrow a+b+2016=2017\)

\(\Leftrightarrow a+b=1\)       \(\left(1\right)\)

\(F\left(-1\right)=2018\)

\(\Leftrightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=2018\)

\(\Leftrightarrow a-b+c=2018\)

\(\Leftrightarrow a-b+2016=2018\)

\(\Leftrightarrow a-b=2\)       \(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=\left(1+2\right)\div2=3\div2=1.5\)

\(\Rightarrow b=1-1.5=-0.5\)

Vậy \(F\left(x\right)=1.5x^2-0.5x+2016\)

\(\Leftrightarrow F\left(2\right)=1.5\cdot2^2-0.5\cdot2+2016\)

\(=1.5\cdot4-0.5\cdot2+2016\)

\(=6-1+2016=2021\)

Vậy \(F\left(2\right)=2021\)

nhớ k nha

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

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

\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (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) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)

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

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

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

Quá i dì

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

\(\Rightarrow c=2010\)

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

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

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

\(\Rightarrow a-b+2010=2012\Rightarrow a-b=2\)(2)

Từ (1) và (2) suy ra \(\hept{\begin{cases}a=\frac{1+2}{2}=\frac{3}{2}\\b=\frac{1-2}{2}=\frac{-1}{2}\end{cases}}\)

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

\(\Rightarrow f\left(-2\right)=\frac{3}{2}.4+\frac{1}{2}.2+2010=2017\)

- Ta có: \(f\left(x\right)=a.x^2+b.x+c\)

 + \(f\left(0\right)=a.0^2+b.0+c=c=2010\)   (1)

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

 + \(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c=2012\)   (3)

- Thay \(c=2010\)vào đa thức (2), (3), ta có:

   \(\hept{\begin{cases}a+b=2011-2010\\a-b=2012-2010\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=1\\a-b=2\end{cases}}\)

- Ta lại có: \(a-b=2\)

          \(\Leftrightarrow a=b+2\)

- Thay \(a=b+2\)vào đa thức: \(a+b=1\), ta có:

       \(b+2+b=1\)

\(\Leftrightarrow2b=-1\)

\(\Leftrightarrow b=-\frac{1}{2}=-0,5\)

- Thay \(b=-0,5\)vào đa thức: \(a+b=1\), ta có:

       \(a-0,5=1\)

\(\Leftrightarrow a=1,5\)

Vậy hàm số \(y=f\left(x\right)\)có dạng: \(y=f\left(x\right)=1,5x^2-0,5x+2010\)

\(\Rightarrow f\left(-2\right)=1,5.\left(-2\right)^2-0,5.\left(-2\right)+2010=2017\)

Vậy \(f\left(-2\right)=2017\)

\(f\left(0\right)=c=2010\)

\(f\left(1\right)=a+b+2010=2011\Rightarrow a+b=1\)(1)

\(f\left(-1\right)=a-b+2010=2012\Rightarrow a-b=2\)(2)

Từ (1) và (2) => a = 3/2; b = -1/2.

Vậy \(f\left(-2\right)=\frac{3}{2}\left(-2\right)^2-\frac{1}{2}\left(-2\right)+2010=6+1+2010=2017\)

Xét đa thức \(F\left(x\right)=ax^2+bx+c\)

\(F\left(0\right)=c=2016\)

\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\)  (1)

\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\)  (2)

Từ (1), (2)

\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)

\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)

Vậy \(F\left(2\right)=2021\).