Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-3m\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x_3=\dfrac{2}{x_1^2}\\x_4=\dfrac{2}{x^2_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2}{x_1^2}+\dfrac{2}{x_2^2}\\x_3x_4=\dfrac{4}{x_1^2x_2^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2\left(x_1+x_2\right)^2-4x_1x_2}{\left(x_1x_2\right)^2}\\x_3x_4=\dfrac{4}{\left(x_1x_2\right)^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2.\left(-5\right)^2-4\left(-3m\right)}{\left(-3m\right)^2}=\dfrac{12m+50}{9m^2}\\x_3x_4=\dfrac{4}{\left(-3m\right)^2}=\dfrac{4}{9m^2}\end{matrix}\right.\)
\(\Rightarrow x_3;x_4\) là nghiệm:
\(x^2-\left(\dfrac{12m+50}{9m^2}\right)x+\dfrac{4}{9m^2}=0\)
\(\Leftrightarrow9m^2x^2-\left(12m+50\right)x+4=0\)
