\(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)

\(\Rightarrow f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)

\(\Rightarrow f\left(-2\right)+f\left(3\right)=4a-2b+c+9a+3b+c=13a+b+2c=0\)

\(\Rightarrow f\left(-2\right)+f\left(3\right)=0\Rightarrow f\left(-2\right)=-f\left(3\right)\)

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?

Ta có:

f(−2)+f(3)=((−2)2a−2b+c)+(32a+3b+c)=(4a−2b+c)+(9a+3b+c)=13a+b+2c=0f(−2)+f(3)=((−2)2a−2b+c)+(32a+3b+c)=(4a−2b+c)+(9a+3b+c)=13a+b+2c=0

Suy ra⎡⎢ ⎢ ⎢ ⎢⎣{f(−2)>0f(3)<0{f(−2)<0f(3)>0⇒f(−2).f(3)<0

vậy......

\(13a+b+2c=0\Rightarrow b=-13a-2c\)

\(f\left(x\right)=ax^2+bx+c\)

\(f\left(-2\right).f\left(3\right)=\left(4a-2b+c\right)\left(9a+3b+c\right)\)

\(=\left(4a-2\left(-13a-2c\right)+c\right)\left(9a+3\left(-13a-2c\right)+c\right)\)

\(=\left(4a+26a+4c+c\right)\left(9a-39a-6c+c\right)\)

\(=\left(30a+5c\right)\left(-30a-5c\right)\)

\(=-\left(30a+5c\right)^2\le0\)

-Dấu "=" xảy ra khi \(a=-b=-\dfrac{1}{6}c\)

dấu bằng xảy ra khi a = -b = -1/6c

thay f-2 và f3 vào rồi pạn sẽ tìm ra

\(\left\{{}\begin{matrix}f\left(-2\right)=4a-2b+c\\f\left(3\right)=9a+3b+c\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-b=13a+2c\\f\left(-2\right)=30a+5c\\f\left(3\right)=-30a-5c\end{matrix}\right.\) \(\Rightarrow f\left(-2\right).f\left(3\right)=-\left(30a+5c\right)^2\le0\Rightarrow dpcm\)

cộng f(-2)+f(3)=0(gt)

vậy hai số f(-2) và f(3) là hai số đối nhau hoặc bằng không. thế là ra rồi đấy

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