Question

Answer 1

`1.` 

Xét \(\Delta'=\left(m-1\right)^2+4-2m\\ =m^2-2m+1+4-2m\\ =m^2-4m+4+1\\ =\left(m-2\right)^2+1\ge1>0\forall m\in R\)

`=>` Pt có `2` nghiệm phân biệt 

`2.` 

Theo Vi-ét ta có :

\(x_1+x_2=2m-2\\ x_1\cdot x_2=2m-4\\ x^2_1+x_2^2=\left(x_1+x_2\right)^2-2x_1\cdot x_2\\ =\left(2m-2\right)^2-2\left(2m-4\right)\\ =4m^2-8m+4-4m+8\\ =4m^2-12m+12\\ =\left(2m-3\right)^2+3\ge3\)

Vậy \(A_{Min}=x_1^2+x_1^2=3\) Khi `m=3/2`