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\(f\left(x\right)=ax^2+bx+c\)
\(f\left(2\right)=4a+2b+c\)
\(f\left(-1\right)=a-b+c\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=4a+2b+c+a-b+c\)
\(\Leftrightarrow f\left(2\right)+f\left(-1\right)=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=0\Leftrightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Leftrightarrow f\left(2\right).f\left(-1\right)=-f\left(-1\right).f\left(-1\right)\le0\)
\(\Rightarrowđpcm\)
f(0)=a0+b0+c=2010
=>c=2010
f(1)=a1+b1+c=a1+b1+2010
=>a+b=1 (1)
f(-1)=a1+(-b1)+c=a1-b1+2010
=>a-b=2 (2)
Từ (1) và (2) => a=(2+1):2=1,5
b=(1-2):2=-0,5
Vậy f(2)=1,5.2+(-0,5)x2+2010=2014
a) \(\hept{\begin{cases}f\left(2\right)=156\\f\left(-3\right)=156\\f\left(-1\right)=132\end{cases}\Rightarrow\hept{\begin{cases}4a+2b+c=156\\9a-3b+c=156\\a-b+c=132\end{cases}\Rightarrow}\hept{\begin{cases}4a+2b+132-a+b=156\\9a-3b+132-a+b=156\\c=132-a+b\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a+3b=24\\8a-2b=24\\c=132-a+b\end{cases}\Rightarrow\hept{\begin{cases}a+b=8\\-4a+b=-12\\c=132-a+b\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}5a=20\\b=8-a\\c=132-a+b\end{cases}\Rightarrow\hept{\begin{cases}a=4\\b=4\\c=132\end{cases}}}\)
b) \(f\left(x\right)=4x^2+4x+132=4x^2+2x+2x+1+131=2x\left(2x+1\right)+\left(2x+1\right)+131\)
\(=\left(2x+1\right)^2+131\)
\(\left(2x+1\right)^2\ge0\forall x\Rightarrow f\left(x\right)\ge131\forall x\). Vậy \(f\left(x\right)\ne0\forall x\)
\(\left\{{}\begin{matrix}f\left(0\right)=2014\Rightarrow c=2014\left(1\right)\\f\left(1\right)=2015\Rightarrow a+b+c=2015\left(2\right)\\f\left(-1\right)=2017\Rightarrow a-b+c=2017\left(3\right)\end{matrix}\right.\)
\(f\left(-2\right)=4a-2b+c\)
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 \(f\left(0\right)=1\)
\(\Rightarrow a\cdot0^2+b\cdot0+c=1\\ \Rightarrow0+0+c=1\\ \Rightarrow c=1\)
\(f\left(1\right)=0\\ \Rightarrow a\cdot1^2+b\cdot1+c=0\\ \Rightarrow a+b+c=0\\ \Rightarrow a+b=-1\left(1\right)\)
\(f\left(-1\right)=6\\ \Rightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=6\\ \Rightarrow a-b+c=6\\ \Rightarrow a-b=5\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow2a=4\\ \Rightarrow a=2\\ \Rightarrow b=-1-a=-1-2=-3\)
Vậy a = 2 ; b = -3 ; c = 1
\(f\left(x\right)=ax^2+bx+c\)
+ \(f\left(0\right)=1.\)
\(\Rightarrow f\left(0\right)=a.0^2+b.0+c=1\)
\(\Rightarrow f\left(0\right)=a.0+b.0+c=1\)
\(\Rightarrow f\left(0\right)=0+0+c=1\)
\(\Rightarrow f\left(0\right)=c=1\)
\(\Rightarrow c=1.\)
+ \(f\left(1\right)=0.\)
\(\Rightarrow f\left(1\right)=a.1^2+b.1+c=0\)
\(\Rightarrow f\left(1\right)=a.1+b.1+c=0\)
\(\Rightarrow f\left(1\right)=a+b+c=0\)
\(\Rightarrow a+b+c=0\)
Mà \(c=1\left(cmt\right).\)
\(\Rightarrow a+b+1=0\)
\(\Rightarrow a+b=0-1\)
\(\Rightarrow a+b=-1\) (1).
+ \(f\left(-1\right)=6.\)
\(\Rightarrow f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a.1+b.\left(-1\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a+\left(-b\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a-b+c=6\)
\(\Rightarrow a-b+c=6\)
Mà \(c=1\left(cmt\right).\)
\(\Rightarrow a-b+1=6\)
\(\Rightarrow a-b=6-1\)
\(\Rightarrow a-b=5\) (2).
Cộng theo vế (1) và (2) ta được:
\(a+b+a-b=\left(-1\right)+5\)
\(\Rightarrow2a=4\)
\(\Rightarrow a=4:2\)
\(\Rightarrow a=2.\)
+ Ta có: \(a+b=-1.\)
\(\Rightarrow2+b=-1\)
\(\Rightarrow b=\left(-1\right)-2\)
\(\Rightarrow b=-3.\)
Vậy \(a=2;b=-3;c=1.\)
Chúc bạn học tốt!
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(x-1\right)=a\left(x-1\right)^2+b\left(x-1\right)+c\)
\(\Rightarrow f\left(x\right)-f\left(x-1\right)=ax^2+bx+c-ax^2+2ax-a-bx+b-c=x\)
\(\Leftrightarrow2ax-a+b-x=0\)
\(\Leftrightarrow\left(2a-1\right)x+b-a=0\)
\(\Leftrightarrow\hept{\begin{cases}2a-1=0\\b-a=0\end{cases}\Leftrightarrow}a=b=\frac{1}{2}\)
\(\)và Hàm số đúng với mọi giá trị của \(c\)
Vậy \(a=b=\frac{1}{2};c\in R\)