Question

q(x)=ax^2+bx+c chứngminh nếu 5a-b+c=0 thì q(-3).Q(1)< hoặc =0

Answer 1

Ta có ; \(Q_{\left(-3\right)}=9a-3b+c\\ Q_{\left(1\right)}=a+b+c\)

Lại có : \(Q_{\left(-3\right)}+Q_{\left(1\right)}=9a-3b+c+a+b+c=10a-2b+2c\\ =2\left(5a-b+c\right)\)

Mà 5a -b + c = 0 \(\Rightarrow Q_{\left(-3\right)}+Q_{\left(1\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}Q_{\left(-3\right)}\ge0\\Q_{\left(1\right)}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}Q_{\left(-3\right)}< 0\\Q_{\left(1\right)}\ge0\end{matrix}\right.\end{matrix}\right.\Rightarrow Q_{\left(-3\right)}\cdot Q_{\left(1\right)}\ge0\)

\(\RightarrowĐpcm\)