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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 \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)
\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)
\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (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) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)
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 ....
\(f\left(0\right)=c=2010\)
\(f\left(1\right)=a+b+2010=2011\Rightarrow a+b=1\)(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 \(f\left(-2\right)=\frac{3}{2}\left(-2\right)^2-\frac{1}{2}\left(-2\right)+2010=6+1+2010=2017\)
\(f\left(0\right)=2010\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a+b+2010=2011\Rightarrow a+b=1\left(1\right)\)
\(f\left(-1\right)=2012\Rightarrow a-b+2010=2012\Rightarrow a-b=2\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\)a=1,5 ; b= \(-0,5\)
\(\Rightarrow f\left(-2\right)=1,5\times\left(-2\right)^2+\left(-0,5\right)\times\left(-2\right)+2010=2005\)
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 \(\hept{\begin{cases}a=\frac{1+2}{2}=\frac{3}{2}\\b=\frac{1-2}{2}=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow f\left(x\right)=\frac{3}{2}x^2-\frac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\frac{3}{2}.4+\frac{1}{2}.2+2010=2017\)