Ta có: \(\Delta=\left\lbrack2\left(m-3\right)\right\rbrack^2-4\left(3m^2-8m+5\right)\)
\(=4\left(m^2-6m+9\right)-12m^2+32m-20\)
\(=4m^2-24m+36-12m^2+32m-20=-8m^2+8m+16\)
\(=-8\left(m^2-m-2\right)=-8\left(m-2\right)\left(m+1\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>-8(m-2)(m+1)>=0
=>(m-2)(m+1)<=0
=>-1<=m<=2
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2\left(m-3\right)\\ x_1x_2=\frac{c}{a}=3m^2-8m+5=\left(3m-5\right)\left(m-1\right)\end{cases}\)
\(x_1^2+2x_2^2-3x_1x_2=x_1-x_2\)
=>\(\left(x_1-x_2\right)\left(x_1-2x_2\right)-\left(x_1-x_2\right)=0\)
=>\(\left(x_1-x_2\right)\left(x_1-2x_2-1\right)=0\)
TH1: \(x_1-x_2=0\)
=>\(x_1=x_2\)
mà \(x_1+x_2=2\left(m-3\right)\)
nên \(x_1=x_2=\frac{2\left(m-3\right)}{2}=m-3\)
\(x_1x_2=3m^2-8m+5\)
=>\(3m^2-8m+5=\left(m-3\right)^2=m^2-6m+9\)
=>\(2m^2-2m-4=0\)
=>\(m^2-m-2=0\)
=>(m-2)(m+1)=0
=>\(\left[\begin{array}{l}m-2=0\\ m+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=2\left(nhận\right)\\ m=-1\left(nhận\right)\end{array}\right.\)
TH2: \(x_1-2x_2-1=0\)
=>\(x_1-2x_2=1\)
mà \(x_1+x_2=2\left(m-3\right)=2m-6\)
nên \(x_1-2x_2-x_1-x_2=1-2m+6=-2m+7\)
=>\(-3x_2=-2m+7\)
=>\(x_2=\frac{2m-7}{3}\)
\(x_1+x_2=2m-6\)
=>\(x_1=2m-6-\frac{2m-7}{3}=\frac{3\left(2m-6\right)-2m+7}{3}=\frac{4m-11}{3}\)
\(x_1x_2=3m^2-8m+5\)
=>\(\frac{\left(2m-7\right)\left(4m-11\right)}{9}=3m^2-8m+5\)
=>\(9\left(3m^2-8m+5\right)=\left(2m-7\right)\left(4m-11\right)\)
=>\(27m^2-72m+45=8m^2-50m+77\)
=>\(19m^2-22m-32=0\)
=>(19m+16)(m-2)=0
=>\(\left[\begin{array}{l}19m+16=0\\ m-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=-\frac{16}{19}\left(nhận\right)\\ m=2\left(nhận\right)\end{array}\right.\)
