Để pt có 2 nghiệm pb khác 0
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'=1-\left(m-5\right)>0\\m-5\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< 6\\m\ne5\end{matrix}\right.\)
Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-5\end{matrix}\right.\)
\(\frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{10}{9}\Leftrightarrow\frac{x_1^2+x_2^2}{x_1^2x_2^2}=\frac{10}{9}\)
\(\Leftrightarrow9\left(x_1^2+x_2^2\right)=10\left(x_1x_2\right)^2\)
\(\Leftrightarrow9\left(x_1+x_2\right)^2-18x_1x_2=10\left(x_1x_2\right)^2\)
\(\Leftrightarrow36-18\left(m-5\right)=10\left(m-5\right)^2\)
\(\Leftrightarrow5\left(m-5\right)^2+9\left(m-5\right)-18=0\)
\(\Rightarrow\left[{}\begin{matrix}m-5=-3\\m-5=\frac{6}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=2\\m=\frac{31}{5}>6\left(l\right)\end{matrix}\right.\)