\(\Delta=\left(2m-2\right)^2-4\left(m^2-4m+3\right)\)

\(=4m^2-8m+4-4m^2+16m-12\)

=8m-8

Để phương trình có hai nghiệm thì 8m-8>=0

=>8m>=8

=>m>=1

Theo Vi-et, ta có:

\(x_1+x_2=-\frac{b}{a}=2\left(m-1\right)=2m-2\)

\(x_1x_2=\frac{c}{a}=m^2-4m+3\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=\left(2m-2\right)^2-4\left(m^2-4m+3\right)\)

\(=4m^2-8m+4-4m^2+16m-12=8m-8\)

x2 là nghiệm của phương trình

=>\(x_2^2-2\left(m-1\right)\cdot x_2+m^2-4m+3=0\)

=>\(x_2^2-2m\cdot x_2+2\cdot x_2+m^2-4m+3\)

=>\(2m\cdot x_2=x_2^2+2x_2+m^2-4m+3\)

\(2mx_2+x_1^2+2x_1+x_1x_2+1\)

\(=x_1^2+x_2^2+2x_1+2x_2+m^2-4m+3+1+x_1x_2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)+x_1x_2+m^2-4m+4\)

\(=\left(2m-2\right)^2+2\left(2m-2\right)-\left(m^2-4m+3\right)+m^2-4m+4\)

\(=4m^2-8m+4+4m-4-m^2+4m-3+m^2-4m+4=4m^2-4m+1=\left(2m-1\right)^2\)

Ta có: \(A=m^2+20-\left(x_1-x_2\right)^2-\sqrt{2mx_2+x_1^2+2x_1+x_1x_2+1}\)

\(=m^2+20-8m+8-\sqrt{\left(2m-1\right)^2}=m^2-8m+28-\left|2m-1\right|\)

Ta có:m>=1

=>2m>=2

=>2m-1>=1>0

=>\(A=m^2-8m+28-\left(2m-1\right)=m^2-10m+29\)

\(=m^2-10m+25+4=\left(m-5\right)^2+4\ge4\forall m\)

Dấu '=' xảy ra khi m-5=0

=>m=5