Để pt có 2 nghiệm
\(\left\{{}\begin{matrix}2m-1\ne0\\\Delta'=\left(m+4\right)^2-\left(2m-1\right)\left(5m+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne\frac{1}{2}\\-9x^2+9x+18\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{1}{2}\\-1\le x\le2\end{matrix}\right.\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2\left(m+4\right)}{2m-1}\\x_1x_2=\frac{5m+2}{2m-1}\end{matrix}\right.\)
\(x_1^2+x_2^2-2x_1x_2-16=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-16=0\)
\(\Leftrightarrow\frac{\left(m+4\right)^2}{\left(2m-1\right)^2}-\frac{\left(5m+2\right)}{2m-1}-4=0\)
\(\Leftrightarrow\left(m+4\right)^2-\left(5m+2\right)\left(2m-1\right)-4\left(2m-1\right)^2=0\)
\(\Leftrightarrow-25m^2+25m+14=0\Rightarrow\left[{}\begin{matrix}m=\frac{7}{5}\\m=-\frac{2}{5}\end{matrix}\right.\) (nhận)
