\(Δ'=[-(m+1)]^2-1.(m^2+m-1)=m^2+2m+1-m^2-m+1=m+2\)
Pt có hai nghiệm
\(→Δ'=m+2\ge 0\\↔m\ne -2\)
Theo hệ thức Vi-ét:
\(\begin{cases}x_1+x_2=\dfrac{-b}{a}=2(m+1)\\x_1x_2=\dfrac c a=m^2+m-1\end{cases}\)
\(-x_1^2-x_2^2\\=-(x_1^2+x_2^2)\\=-(x_1^2+2x_1x_2+x_2^2-2x_1x_2)\\=-(x_1+x_2)^2+2x_1x_2\\=-[2(m+1)]^2+2(m^2+m-1)\\=-(4m^2+8m+4)+2m^2+2m-2\\=-4m^2-8m-4+2m^2+2m-2\\=-2m^2-6m-6\\=-2(m^2+3m+3)\\=-2(m^2+2.\dfrac{3}{2}m+\dfrac{9}{4}+\dfrac{3}{4})\\=-2(m+\dfrac{3}{2})^2-\dfrac{3}{2}\)
Vì \(-2(m+\dfrac{3}{2})^2\le 0→-2(m+\dfrac{3}{2})^2-\dfrac{3}{2}\le -\dfrac{3}{2}\\→\max=-\dfrac{3}{2}\)
\(→\) Dấu "=" xảy ra khi \(m+\dfrac{3}{2}=0\)
\(↔m=-\dfrac{3}{2}\)
Vậy \(\min -x_1^2-x_2^2=-\dfrac{3}{2}\) khi \(m=-\dfrac{3}{2}\)
