\(\Delta=b^2-4ac=\left(-2m\right)^2-4\cdot1\cdot\left(m+2\right)=4m^2-4m-8>0\forall m\)
=> ptr luôn có 2 nghiệm phân biệt với mọi m.
Theo định lý Viete, ta có:
\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=2m\\P=x_1x_2=\dfrac{c}{a}=m+2\end{matrix}\right.\)
Ta có: \(x_1^3+x_2^3=16\)
\(\Leftrightarrow\left(x_1+x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=16\)
\(\Leftrightarrow\left(x_1+x_2\right)\left[\left(x_1^2+x_2^2\right)-x_1x_2\right]=16\)
\(\Leftrightarrow S\left(S^2-2P-P\right)=16\)
\(\Leftrightarrow S^3-3PS=16\)
\(\Leftrightarrow2m^3-3\cdot2m\cdot\left(m+2\right)=16\)
\(\Leftrightarrow8m^3-6m^2-12m-16=0\)
\(\Leftrightarrow m=2\left(bam:Mode-5-4\right)\)
Vậy với m = 2 thì tmycđb
