<=>\(\hept{\begin{cases}4x^2+2mx=2\\mx^2-x=-2\end{cases}}\)<=>\(\hept{\begin{cases}\left(4+m\right)x^2+\left(2m-1\right)x=0\\mx^2-x=-2\end{cases}}\)<=>\(\hept{\begin{cases}x\left(\left(m+4\right)x+2m-1\right)=0\\mx^2-x=-2\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\mx^2-x=-2\end{cases}}\)(vô nghiệm) hoặc \(\hept{\begin{cases}x=\frac{1-2m}{m+4}\\mx^2-x=-2\end{cases}}\)(điều kiện m\(\ne-4\)) <=>m(\(\frac{1-2m}{m+4}\))2-\(\frac{1-2m}{m+4}\)=-2 <=> m(1-2m)2-(1-2m)(m+4)=-2(m+4)2 <=> 4m3-4m2+m-m+2m2-4+8m=-2m2-16m-32 <=> 4m3+24m+28=0
<=> (m+1)(4m2-4m+28)=0 <=>m+1=0 (vì 4m2-4m+28=(2m-1)2+27>0) <=> m=-1 (thỏa mãn m\(\ne-4\))
Vậy m=-1