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