\(x^3-3mx^2+6m-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-3mx\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-\left(3m-2\right)x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\f\left(x\right)=x^2-\left(3m-2\right)x+4=0\left(1\right)\end{matrix}\right.\)
Xét (1), \(\Delta=\left(3m-2\right)^2-16>0\Rightarrow\left[{}\begin{matrix}m>2\\m< -\dfrac{2}{3}\end{matrix}\right.\)
Khi đó: \(\left\{{}\begin{matrix}x_1+x_2=3m-2\\x_1x_2=4\end{matrix}\right.\)
TH1: \(x_1< x_2< 2\Rightarrow2x_1=x_2^2\Rightarrow\dfrac{8}{x_2}=x_2^2\Rightarrow x_2=2\) (ktm pt có 3 nghiệm pb)
TH2: \(2< x_1< x_2\Rightarrow2x_2=x_1^2\) tương tự như trên \(\Rightarrow x_1=2\) (ktm)
TH3: \(x_1< 2< x_2\Rightarrow x_1x_2=4\) (luôn đúng)
Để xảy ra điều này thì: \(f\left(2\right)=2^2-\left(3m-2\right).2+4< 0\Rightarrow m>2\)
\(\Rightarrow m=\left\{3;4;5\right\}\)
