Để phương trình có nghiệm, $\Delta'\ge 0$
$\Delta'=(m+4)^2-m^2+8$
$=m^2+8m+8-m^2+8$
$=8m+16\ge 0$
$\Leftrightarrow m\ge -2$
Theo Viet: $x_1+x_2=2(m+4)=2m+8; x_1x_2=m^2-8$
a,
$A=x_1^2+x_2^2-x_1-x_2$
$=(x_1+x_2)^2-(x_1+x_2)-2x_1x_2$
$=(2m+8)^2-2m-8-2(m^2-8)$
$=4m^2+32m+64-2m-8-2m^2+16$
$=2m^2+30m+72$
$=2(m^2+15m+36)$
$=2(m^2+2.m.7,5+56,25-20,25)$
$=2(m+7,5)^2-\dfrac{81}{2}\ge \dfrac{81}{2}$
$\to A_{\min}=\dfrac{-81}{2}\Leftrightarrow m=-7,5$ (Loại)
Vậy không có $m$ để $A_{\min}$
b,
$B=x_1+x_2-3x_1x_2$
$=2m+8-3(m^2-8)$
$=-3m^2+2m+32$
$=-(3m^2-2m-32)$
$=-\Bigg[ (\sqrt3 m)^2-2.\sqrt3 m.\dfrac{1}{\sqrt3}+\dfrac{1}{3}-\dfrac{97}{3}\Bigg]$
$=-\Bigg(\sqrt3 m-\dfrac{1}{\sqrt3}\Bigg)^2+\dfrac{97}{3}\le \dfrac{97}{3}$
$\to B_{\max}=\dfrac{97}{3}\Leftrightarrow m=\dfrac{1}{3}$ (TM)
$C=x_1^2+x_2^2-x_1x_2$
$=(x_1+x_2)^2-3x_1x_2$
$=(2m+8)^2-3(m^2-8)$
$= 4m^2+32m+64-3m^2+24$
$=m^2+32m+ 88$
$\to$ không tồn tại $m$ để $C_{\max}$
