\(\Delta=\left(m+2\right)^2-4\cdot\left(-1\right)\cdot\left(-2m+5\right)\)
\(=m^2+4m+4-8m+20\)
\(=m^2-4m+24\)
\(=\left(m-2\right)^2+20>0\)
Vậy: Phương trình luôn có hai nghiệm phân biệt
Theo đề, ta có:
\(A=\left(x_1+x_2\right)^2-6x_1x_2+7\)
\(=\left(m+2\right)^2-6\cdot\dfrac{-2m+5}{-1}+7\)
\(=m^2+4m+4+6\left(-2m+5\right)+7\)
\(=m^2+4m+11-12m+30\)
\(=m^2-8m+41\)
\(=\left(m-4\right)^2+25>=25\forall m\)
Dấu '=' xảy ra khi m=4
\(\Delta=\left(m+2\right)^2+4\left(-2m+5\right)=m^2-4m+24=\left(m-2\right)^2+20>0\)
Vậy pt luôn có 2 nghiệm pb
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=2m-5\end{matrix}\right.\)
\(A=\left(x_1+x_2\right)^2-6x_1x_2+7\)
\(=\left(m+2\right)^2-6\left(2m-5\right)+7=m^2-8m+34\)
\(=m^2-8m+16+18=\left(m-4\right)^2+18\ge18\forall m\)
Dấu ''='' xảy ra khi m = 4
Vậy m = 4 thì A đặt GTNN là 18
