Question

cho đt f(x)=x²+ax+b,biết rầng a+\(\dfrac{b}{3}\)=-4.tìm 1 số c biết f(c)=-3

Answer 1

Từ \(a+\dfrac{b}{3}=-4\Rightarrow\dfrac{b}{3}=-4-a\) (1)

Ta có: \(f\left(c\right)=c^2+ac+b=-3\)

\(\Leftrightarrow c^2+ac+b+3=0\)

\(\Leftrightarrow c^2+ac+3\left(\dfrac{b}{3}+1\right)=0\)

\(\Leftrightarrow c^2+ac+3\left(-4-a+1\right)=0\) (theo 1)

\(\Leftrightarrow c^2+ac-3a-9=0\)

\(\Leftrightarrow\left(c^2-9\right)+\left(ac-3a\right)=0\)

\(\Leftrightarrow\left(c-3\right)\left(c+3\right)+a\left(c-3\right)=0\)

\(\Leftrightarrow\left(c-3\right)\left(c+3+a\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}c-3=0\\c+3+a=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=3\\a=-6\end{matrix}\right.\)

Vậy \(c=3\)

Answer 2

ai hiu dc bài này mời nhận giải fields

Answer 3

Ace LegonaNguyễn Huy Tú