\(y'=x^2-2\left(m-2\right)x+1\)
Hàm có 2 cực trị khi \(\Delta'=\left(m-2\right)^2-1>0\Rightarrow\left[{}\begin{matrix}m>3\\m< 1\end{matrix}\right.\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=1\end{matrix}\right.\)
\(x_1+\dfrac{2}{x_2}=\dfrac{3}{x_1}+m\Leftrightarrow x_1+2x_1=3x_2+m\)
\(\Leftrightarrow x_1-x_2=\dfrac{m}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2m-4\\x_1-x_2=\dfrac{m}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{7m}{6}-2\\x_2=\dfrac{5m}{6}-2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{7m}{6}-2\right)\left(\dfrac{5m}{6}-2\right)=1\)
\(\Leftrightarrow35m^2-144m+108=0\)
\(\Rightarrow m_1+m_2=\dfrac{144}{35}\approx4,11\)
Chắc là gần A nhất?
