\(ac=-1< 0\Rightarrow\) pt luôn có 2 nghiệm phân biệt
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-1\end{matrix}\right.\)
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}+\frac{10}{3}=0\Leftrightarrow\frac{x_1^2+x_2^2}{x_1x_2}+\frac{10}{3}=0\)
\(\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}+\frac{10}{3}=0\Leftrightarrow\frac{4\left(m-1\right)^2+2}{-1}+\frac{10}{3}=0\)
\(\Leftrightarrow4m^2-8m+\frac{8}{3}=0\Rightarrow\left[{}\begin{matrix}m=\frac{3+\sqrt{3}}{3}\\m=\frac{3-\sqrt{3}}{3}\end{matrix}\right.\)