Đặt \(\left\{{}\begin{matrix}u=x^2-2x+m\\v=x^2+2\end{matrix}\right.\) \(\Rightarrow f'\left(x\right)=\frac{u'v-uv'}{v^2}=0\)
\(\Leftrightarrow u'v=uv'\Leftrightarrow\frac{u}{v}=\frac{u'}{v'}\)
\(\Rightarrow f\left(x_1\right)=\frac{u\left(x_1\right)}{v\left(x_1\right)}=\frac{u'\left(x_1\right)}{v'\left(x_1\right)}=\frac{2x_1-2}{2x_1}=1-\frac{1}{x_1}\)
\(f\left(x_2\right)=\frac{u'\left(x_2\right)}{v'\left(x_2\right)}=\frac{2x_2-2}{2x_2}=1-\frac{1}{x_2}\)
\(\Rightarrow k=\frac{1-\frac{1}{x_1}-1+\frac{1}{x_2}}{x_1-x_2}=\frac{1}{x_1x_2}\)
Mặt khác \(x_1;x_2\) là nghiệm của
\(f'\left(x\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+2\right)-2x\left(x^2-2x+m\right)=2x^2-2\left(m-2\right)x-4=0\)
\(\Rightarrow x_1x_2=-\frac{4}{2}=-2\)
\(\Rightarrow k=-\frac{1}{2}\)
