\(\Delta'=1-m+5=6-m>0\Rightarrow m< 6\)
Khi đó theo Viet pt có 2 nghiệm: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-5\end{matrix}\right.\)
Để biểu thức đề bài xác định \(\Rightarrow\left\{{}\begin{matrix}x_1\ne0\\x_2\ne0\end{matrix}\right.\) \(\Rightarrow m\ne5\)
Ta có:
\(\frac{1}{x_1^2}+\frac{1}{x^2_2}=\frac{10}{9}\Leftrightarrow\frac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\frac{10}{9}\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow9\left(x_1+x_2\right)^2-18x_1x_2-10\left(x_1x_2\right)^2=0\)
\(\Leftrightarrow36-18\left(m-5\right)-10\left(m-5\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}m-5=\frac{6}{5}\\m-5=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\frac{31}{5}>6\left(l\right)\\m=2\end{matrix}\right.\)
