\(\Delta=\left(2m-2\right)^2-4\cdot1\left(m^2+3m-5\right)\)
\(=4m^2-8m+4-4m^2-12m+20=-20m+24\)
Để phương trình có hai nghiệm thì Δ>=0
=>-20m+24>=0
=>-20m>=-24
=>m<=1,2
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2\left(m-1\right);x_1x_2=\frac{c}{a}=m^2+3m-5\)
\(x_1^2-2\left(m-1\right)x_1+m^2+3m-5=0\)
=>\(x_1^2+2x_1+1-2m\cdot x_1+m^2+3m-6=0\)
=>\(\left(x_1+1\right)^2=2mx_1-m^2-3m+6\)
\(\left(x_1+1\right)^2+2m\cdot x_2-2=0\)
=>\(2m\left(x_1+x_2\right)-m^2-3m+6-2=0\)
=>\(2m\cdot2\left(m-1\right)-m^2-3m+4=0\)
=>\(4m\left(m-1\right)-m^2-3m+4=0\)
=>\(4m^2-4m-m^2-3m+4=0\)
=>\(3m^2-7m+4=0\)
=>\(3m^2-3m-4m+4=0\)
=>(m-1)(3m-4)=0
=>m=1(nhận) hoặc m=4/3(loại)
