$\vec{n_1}(m;1)\to\vec{u_1}(1;-m)$
$\to |\vec{u_1}|=\sqrt{m^2+1}$
$\vec{n_2}(1;1)\to \vec{u_2}(1;-1)$
$\to |\vec{u_2|}=\sqrt{1+1}=\sqrt2$
$\to \cos(\vec{u_1},\vec{u_2})=\dfrac{ 1.1+m.1}{ \sqrt{2}.\sqrt{m^2+1} }=\dfrac{m+1}{\sqrt{2m^2+2}}$
Ta có $(d_1, d_2)=30^o$
$\to (\vec{u_1},\vec{u_2})=30^o$ hoặc $150^o$
Ta có $\cos^230^o=\cos^2150^o$.
$ \pm\dfrac{\sqrt3}{2}=\dfrac{m+1}{ \sqrt{2m^2+2}}$
$\to m+1=\pm\dfrac{\sqrt3}{2}\sqrt{2m^2+2}$
$\to m^2+2m+1=\dfrac{3}{4}(2m^2+2)$
$\to m=2$ hoặc $m=-1$
