Để phương trình có nghiệm x1;x2 thì :
\(\Delta'=\left(m+4\right)^2-\left(m^2-8\right)\)
\(=\left(m^2+8m+16\right)-m^2+8\)
\(=8m+24\ge0\Leftrightarrow m\ge-3\)
Theo hệ thức Viet,ta có :
\(\left\{{}\begin{matrix}x1+x2=2\left(m+4\right)\\x1.x2=m^2-8\end{matrix}\right.\)
a) \(A=x1^2+x2^2-x1-x2=\left(x1+x2\right)^2-\left(x1+x2\right)-2x1x2=4\left(m+4\right)^2-2\left(m+4\right)-2\left(m^2-8\right)\)
\(A=2m^2+30m+66=0\)
\(A=\left(4m+3\right)^2-\frac{519}{8}\ge-\frac{519}{8}\)
b) \(B=2\left(m+4\right)-3\left(m^2-8\right)\)
\(B=-3m^2+2m+32\)
\(B=\frac{97}{3}-\left(3x-1\right)^2\le\frac{97}{3}\Leftrightarrow x=\frac{1}{3}\)
c) \(C=x1^2+x2^2-x1x2=\left(x1+x2\right)^2-3x1x2\)
\(C=4\left(m+4\right)^2-3\left(m^2-8\right)\)
\(C=-3m^2+4m+28\)
\(C=\frac{88}{3}-\left(3x-2\right)^2\le\frac{88}{3}\Leftrightarrow x=\frac{2}{3}\)
