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\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)
\(\Rightarrow f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=4a-2b+c+9a+3b+c=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=0\Rightarrow f\left(-2\right)=-f\left(3\right)\)
Xét \(f\left(-2\right).f\left(3\right)=\left[-f\left(3\right)\right].f\left(3\right)=-\left[f\left(3\right)\right]^2\le0\)
Vậy \(f\left(-2\right).f\left(3\right)\le0\)
mình không hiểu, sao f(−2).f(3)=[−f(3)].f(3)=−[f(3)]2?
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\hept{\begin{cases}f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\\f\left(3\right)=a.3^2+b.3+c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}f\left(-2\right)=4a-2b+c\\f\left(3\right)=9a+3b+c\end{cases}}\)
Ta có: \(f\left(-2\right)+f\left(3\right)=\left(4a-2b+c\right)+\left(9a+3b+c\right)=13a+b+2c=0\)
Suy ra f(-2) và f(3) là hai số đối nhau.
Vậy \(f\left(-2\right).f\left(3\right)\le0\)(Tích hai số đối nhau bé hơn hoặc bằng 0)
(Dấu '="\(\Leftrightarrow f\left(-2\right)=f\left(3\right)=0\))
Lời giải:
a)
\(f(1)=a.1^2+b.1+c=a+b+c\)
\(f(2)=a.2^2+b.2+c=4a+2b+c\)
b)
\(f(-2)=a(-2)^2+b(-2)+c=4a-2b+c\)
Do đó:
\(f(1)+f(-2)=(a+b+c)+(4a-2b+c)=5a-b+2c=0\)
\(\Rightarrow f(-2)=-f(1)\)
\(\Rightarrow f(1)f(-2)=-f(1)^2\leq 0\)
c)
Với $a=1,b=2,c=3$ thì :
\(f(x)=x^2+2x+3=x(x+1)+(x+1)+2=(x+1)(x+1)+2\)
\(=(x+1)^2+2\)
Vì \((x+1)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow f(x)=(x+1)^2+2\geq 2>0\)
Vậy $f(x)\neq 0$
Do đó $f(x)$ không có nghiệm.
Bài 1:
\(f(x)=ax^2+bx+c\Rightarrow \left\{\begin{matrix} f(-2)=a(-2)^2+b(-2)+c=4a-2b+c\\ f(3)=a.3^2+b.3+c=9a+3b+c\end{matrix}\right.\)
\(\Rightarrow f(-2)+f(3)=(4a-2b+c)+(9a+3b+c)\)
\(=13a+b+2c=0\)
\(\Rightarrow f(-2)=-f(3)\Rightarrow f(-2)f(3)=-f(3)^2\leq 0\) do \(f(3)^2\geq 0\)
Ta có đpcm.
Bài 2:
Thay $x=-3$ ta có:
\(f(-3)=a.(-3)+5=-2\)
\(\Rightarrow a=\frac{7}{3}\)
Vậy $a=\frac{7}{3}$
Lời giải:
Ta có:
\(f(-2)=4a-2b+c\)
\(f(3)=9a+3b+c\)
\(\Rightarrow f(-2)+f(3)=13a+b+2c=0\) (theo giả thiết)
\(\Rightarrow f(-2)=-f(3)\Rightarrow f(-2)(f(3)=-f^2(3)\leq 0\)
Do đó ta có đpcm.
Ta có f(-2).f(3)=(4a-2b+c).(9a+3b+c)
=(4a-2b+c).(13a+b+2c-(4a-2b+c)
Mà 13a+b+2c=0\(\Rightarrow\)f(-2).f(3)=\(-\left[\left\{4a-2b+c\right\}^2\right]\)
Có (4a-2b+c)^2 luôn luôn \(\le\)0
Nên f(-2).f(3)\(\le\)0
Ta có :
f(1) + f(-2) = a + b + c + 4a - 2b + c = 5a - b + 2c = 0
\(\Rightarrow\)f(1) = -f(-2)
Do đó : f(1) . f(-2) = -[f(-2)]2 \(\le\)0
a) Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\\f\left(3\right)=a.3^2+b.3+c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=4a-2b+c\\f\left(3\right)=9a+3b+c\end{matrix}\right.\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=\left(4a-2b+c\right)+\left(9a+3b+c\right)\)
\(=\left(4a+9a\right)+\left(-2b+3b\right)+\left(c+c\right)\)
\(=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(3\right)\right]^2\le0\)
Vậy \(f\left(-2\right).f\left(3\right)\le0\) (Đpcm)
b) Sửa đề:
Biết \(5a+b+2c=0\)
Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=a.2^2+b.2+c=4a+2b+c\\f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\end{matrix}\right.\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=\left(a-b+c\right)+\left(4a+2b+c\right)\)
\(=\left(4a+a\right)+\left(-b+2b\right)+\left(c+c\right)\)
\(=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Rightarrow f\left(2\right).f\left(-1\right)=-\left[f\left(-1\right)\right]^2\le0\)
Vậy \(f\left(2\right).f\left(-1\right)\le0\) (Đpcm)
Hình như đề sai nha bạn phải là 5a+b=-2c mới đúng
Có \(5a+b=-2c\Rightarrow5a+b+2c=0\)
\(f\left(x\right)=ax^2+bc+c\)
\(\Rightarrow f\left(-1\right)=a.\left(-1\right)+b.\left(-1\right)+c=a-b+c\)
\(\Rightarrow f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)
\(\Rightarrow f\left(-1\right)+f\left(2\right)=a-b+c+4a+2b+c=5a+b+2c=0\)
\(\Rightarrow f\left(-1\right)+f\left(2\right)=0\Rightarrow f\left(-1\right)=-f\left(2\right)\)
Xét \(f\left(-1\right).f\left(2\right)=[-f\left(2\right)].f\left(2\right)=-[f\left(2\right)]^2\le0\)
Vậy \(f\left(-1\right).f\left(2\right)\le0\)