Theo bài ra ta có:
\(\hept{\begin{cases}c=2016\\a+b+c=2017\\a-b+c=2018\end{cases}\Leftrightarrow2a+2c=4035\Leftrightarrow2a=4035-2016.2=3}\)
\(\Leftrightarrow a=\frac{3}{2}\)
thay vào ta tính dc b nha
a+b+c=2017
a-b+c=2018
công vế theo vế ta dc
tính a,b,c xong rồi thay x=2 ta dc F(2)
Với x=0
\(\Rightarrow f\left(0\right)=0+0+c=2016\Rightarrow c=2016\)
Với x=1
\(f\left(1\right)=a+b+c=2017\Rightarrow a+b=1\)(1)
Với x=-1
\(f\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\)(2)
Cộng (1) và (2) ta được 2a=3 \(\Rightarrow a=\frac{3}{2}\Rightarrow b=2-\frac{3}{2}=\frac{1}{2}\)
Vậy \(f\left(2\right)=4.\frac{3}{2}+2.\frac{1}{2}+2016=6+1+2016=2023\)
Ta có:
\(f\left(0\right)=2016\)hay \(a.0^2+b.0+c=2016\Rightarrow c=2016^{\left(1\right)}\)
\(f\left(1\right)=2017\)hay \(a.1^2+b.1+2016=2017\Rightarrow a+b=1^{\left(2\right)}\)
\(f\left(-1\right)=2018\)hay \(a.\left(-1\right)^2+b.\left(-1\right)+2016=2018\Rightarrow a-b=2^{\left(3\right)}\)
Từ \(^{\left(1\right),\left(2\right),\left(3\right)}\)suy ra:
\(\hept{\begin{cases}a+b+c=2017\\a-b+c=2018\end{cases}\Rightarrow}\left(a+b+c\right)+\left(a-b+c\right)=4053\)
\(\Rightarrow2a+2c=4053\)
\(2\left(a+c\right)=4053\)
\(a+c=\frac{4053}{2}=2017,5\)
\(a+2016=2017,5\Rightarrow a=1,5\)
Mà \(a+b=1\Rightarrow b=1-a=1-1,5=-0,5\)
Vậy: \(F\left(2\right)=1,5.2^2+\left(-0,5\right).2+2016=6-1+2016=2021\)
P/s: @Khuyển Dạ Xoa bước cộng cuối cùng cậu làm sai rồi nhé.
Ta có:\(\hept{\begin{cases}F\left(0\right)=a\cdot0^2+b\cdot0+c=2016\Rightarrow c=2016\\F\left(1\right)=a\cdot1^2+b\cdot1+c=2017\Rightarrow a+b=1\\F\left(-1\right)=a\cdot\left(-1\right)^2+b\left(-1\right)+c=2018\Rightarrow a-b=2\end{cases}}\Rightarrow a+b+a-b=3\Rightarrow a=\frac{3}{2}\Rightarrow b=-\frac{1}{2}\)Ta có:\(F\left(2\right)=a\cdot2^2+b\cdot2+c=4a+2b+c=4\cdot\frac{3}{2}+2\cdot\left(-\frac{1}{2}\right)+2016=2021\)
