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Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=4a-2b+c\)
\(f\left(3\right)=9a+3b+c\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)(vì 13a+b+2c=0)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(-2\right)\right]^2\le0\)( đpcm)
\(f\left(0\right)=5\\ \Leftrightarrow a\cdot0^2+b\cdot0+c=c=5\\\Rightarrow c=5\\ f\left(1\right)=3\\ \Leftrightarrow a\cdot1^2+b\cdot1+c=a+b+5=3\\ \Leftrightarrow a+b=-2\\ \Leftrightarrow2a+2b=-4\\ f\left(-2\right)=4\\ \Leftrightarrow a\cdot\left(-2\right)^2+b\cdot\left(-2\right)+c=4a-2b+5=4\\ \Leftrightarrow4a-2b=-1\\ 2a+2b+4a-2b=-4+\left(-1\right)\\ \Leftrightarrow6a=-5\\ \Leftrightarrow a=\dfrac{-5}{6}\\ a+b=-2\\ \Leftrightarrow\dfrac{-5}{6}+b=-2\\ \Leftrightarrow b=\dfrac{-7}{6}\)
+f(0)= a.0 +b.0 + c =-3 => c = -3
+f(1) = a.12 +b.1-3 = 0 => a+b =3 (1)
+f(-1) = a(-1)+b(-1) -3 =-10 => a -b = -7 (2)
(1)(2) => a =(-7+3):2= -2
b =3-(-2) = 5
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 \(f\left(-2\right)=2022\)
\(\left\{{}\begin{matrix}f\left(0\right)=2014\Rightarrow c=2014\left(1\right)\\f\left(1\right)=2015\Rightarrow a+b+c=2015\left(2\right)\\f\left(-1\right)=2017\Rightarrow a-b+c=2017\left(3\right)\end{matrix}\right.\)
\(f\left(-2\right)=4a-2b+c\)
Lấy (3) nhân 3 công (2) trừ (1) nhân 2
\(f\left(-2\right)=4a-2b+c=3.2017+2015-3.2014\)
\(f\left(-2\right)=3\left(2017-2014\right)+2015=2024\)
Ta có:+)f(2017)=ax^3 + bx +5=5
x(ax^2 + bx)=0
=>ax^2 + bx=0(do x=-2017)
+)f(-2017)=ax^3 + bx +5
=x(ax^2 +bx)+5
=x.0+5=0+5=5