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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 \(F\left(2\right)=2021\).
Theo đề bài f(0)= 2017 => c= 2017
f(1)= 2018 => a + b + c = 2018 => a + b = 1 (1)
f(-1)= 2019 => a - b + c= 2019 => a - b= 2 (2)
Cộng theo vế của (1) và (2), ta được
2a = 3 => a = 3/2
=>b= -1/2
Vậy a=3/2, b=-1/2, c= 2017. Khi đó f(2)= 6 - 2 + 2017= 2021
Vậy f(2)= 2021
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
\(\left\{{}\begin{matrix}f\left(0\right)=2017\\f\left(1\right)=2018\\f\left(-1\right)=2019\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=2017\\a+b+c=2018\\a-b+c=2019\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\a-b=2\\c=2017\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{2}\\b=-\frac{1}{2}\\c=2017\end{matrix}\right.\)
\(\Rightarrow f\left(2\right)=\frac{3}{2}\cdot2^2-\frac{1}{2}\cdot2+2017\)
\(\Rightarrow f\left(2\right)=6-1+2017=2022\)
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, Theo bài ra ta có \(\hept{\begin{cases}f\left(0\right)=c=0\\f\left(1\right)=a+b+c=2013\\f\left(-1\right)=a-b+c=2012\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=2013\\a-b=2012\end{cases}}\)
Cộng vế với vế \(a+b+a-b=2013+2012\Leftrightarrow2a=4025\Leftrightarrow a=\frac{4025}{2}\)
\(\Rightarrow b=\frac{4025}{2}-2012=\frac{1}{2}\)
Vậy \(a=\frac{4025}{2};b=\frac{1}{2};c=0\)
f(0) = a.02 + b. 0 + c = 2016
<=> c =2016
f (1) = a.12 + b.1 + c =2017
<=> a + b =1 (1)
f ( -1 ) = a (-1)2 + b . (-1) +c =2018
<=> a -b =2 (2)
Từ (1),(2) <=> a = 1,5 ; b = -0,5
=> F(x) = 1,5x2 -0,5 x + 2016
F (2) = 1,5 . 22 -0,5 .2 +2016
= 6 -1 +2016 =2021
Ta có:
\(F\left(0\right)=a.0^2+b.0+c=2016\)
\(\Rightarrow c=2016\)
\(F\left(1\right)=a.1^2+b.1+c=2017\)
\(\Rightarrow a+b=1\)
\(F\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=2018\)
\(\Rightarrow a-b=2\)
Vì a + b =1 và a - b = 2 nên \(\Rightarrow a=\frac{3}{2};b=\frac{-1}{2}\)
Vậy \(F\left(2\right)=\frac{3}{2}.2^2-\left(\frac{-1}{2}\right).2+2016=2023\)
Ta có :
\(f\left(0\right)=a.0^2+b.0+c=c=2015\)
\(f\left(1\right)=a.1^2+b.1+c=a+b+c=2016\)
\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c=2017\)
\(a+b+2015=2016\Rightarrow a+b=1\)
\(a-b+2015=2017\Rightarrow a-b=2\)
Cộng vế với vế ta được :\(\left(a+b\right)+\left(a-b\right)=1+2\)
\(\Leftrightarrow2a=3\Rightarrow a=\frac{3}{2}\)
\(\Rightarrow\frac{3}{2}+b=1\Rightarrow b=1-\frac{3}{2}=-\frac{1}{2}\)
\(\Rightarrow f\left(-2\right)=\frac{3}{2}.\left(-2\right)^2+\left(-\frac{1}{2}\right).\left(-2\right)+2015\)
\(=\frac{3}{2}.4+1+2015\)
\(=6+1+2015\)
\(=2022\)
Vậy \(f\left(-2\right)=2022\)