\(\text{Δ}=\left(4m+2\right)^2-4\left(4m^2-2m+3\right)\)
\(=16m^2+16m+4-16m^2+8m-12\)
=24m-8
Để phương trình có hai nghiệm phân biệt thì 24m-8>0
=>m>1/3
\(\left(x_1-1\right)^2+\left(x_2-1\right)^2+2\left(x_1+x_2-x_1x_2\right)=18\)
=>\(x_1^2-2x_1+1+x_2^2-2x_2+1+2\left(x_1+x_2-x_2x_1\right)=18\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-2\left(x_1+x_2\right)+2+2\left(x_1+x_2-x_1x_2\right)=18\)
=>\(\left(4m+2\right)^2-2\left(4m^2-2m+3\right)-2\left(4m+2\right)+2+2\left(4m+2+4m^2-2m+3\right)=18\)
=>\(16m^2+16m+4-8m^2+4m-6-8m-4+2+2\left(4m^2+2m+5\right)=18\)
=>\(8m^2+12m-4+8m^2+8m+10=18\)
=>16m^2+20m+6-18=0
=>16m^2+20m-12=0
=>4m^2+5m-3=0
hay \(m=\dfrac{-5+\sqrt{73}}{8}\)
