1) \(\Delta\)' = \(\left(m-1\right)^2-\left(m-3\right)\)
= \(m^2-2m+1-m+3=m^2-3m+4\)
= \(m^2-2.\dfrac{3}{2}m+\dfrac{9}{4}-\dfrac{9}{4}+4\) = \(\left(m-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\) \(\ge\dfrac{7}{4}>0\forall m\)
\(\Rightarrow\) phương trình luôn có 2 nghiệm phân biệt
2) áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=m-3\end{matrix}\right.\)
ta có :P = \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
thay \(\Leftrightarrow\) \(\left(2m-2\right)^2-2\left(m-3\right)\)
= \(4m^2-8m+4-2m+6\)
= \(4m^2-10m+10\)
= \(\left(2m\right)^2-2.2m.\dfrac{10}{4}+\dfrac{100}{16}-\dfrac{100}{16}+10\)
= \(\left(2m-\dfrac{10}{4}\right)^2+\dfrac{15}{4}\) \(\ge\) \(\dfrac{15}{4}\)
vậy min P = \(\dfrac{15}{4}\) khi \(\left(2m-\dfrac{10}{4}\right)^2=0\) \(\Leftrightarrow\) \(2m-\dfrac{10}{4}=0\)
\(\Leftrightarrow\) \(2m=\dfrac{10}{4}\) \(\Leftrightarrow\) \(m=\dfrac{10}{8}\)
vậy min P là \(\dfrac{15}{4}\) khi m = \(\dfrac{10}{8}\)
