P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

Theo bài ra ta có : 

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

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

hay \(f\left(3\right).f\left(2\right)\ge0\)

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

Dấu ''='' xảy ra <=> \(a=b=c=0\)( thỏa mãn điều kiện )

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

\(=\left[3\left(a+b\right)+6a+c\right]\left[-2\left(a+b\right)+6a+c\right]\)

\(=\left(6a+c\right)\left(6a+c\right)=\left(6a+c\right)^2\ge0\) (đpcm)

Tks anh zai

bạn có thể dùng \(fx\)viết lại không

đề không rõ

a+b+c = 0 => a+b=-c ; b+c=-a ; c+a=-b

=> (1+a/b).(1+b/c).(1+c/a) = a+b/b . b+c/c . c+a/a = -c/b . (-a)/c . (-b)/a = -abc/abc = -1

k mk nha

Ta có: P(-1).P(-2)=[a.(-1)²+b.(-1)+c].[a.(-2)²+b.(-2)+c]

=(a-b+c).(4a-2b+c)

=[(5a-4a)-(3b-2b)+(2c-c)].(4a-2b+c)

=(5a-4a-3b+2b+2c-c).(4a-2b+c)

=[(5a-3b+2c)-(4a-2b+c)].(4a-2b+c)

Vì 5a-3b+2c=0

=>P(-1).P(-2)=[0-(4a-2b+c)].(4a-2b+c)

=-(4a-2b+c).(2a-2b+c)

=-(4a-2b+c)² 

Vì \(\left(4a-2b+c\right)^2\ge0\)

=>\(-\left(4a-2b+c\right)^2\le0\)

=>\(P\left(-1\right).P\left(-2\right)\le0\)

=>ĐPCM