$\left\{{{3x-y=-m}\atop{9x-m^2y=-3\sqrt{3}}}\right.$
Để hệ phương trình có vô số nghiệm
$⇔\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}$
$→\frac{3}{9}=\frac{-1}{-m^2}=\frac{-m}{-3\sqrt{3}}$
$→\frac{1}{3}=\frac{1}{m^2}=\frac{m}{3\sqrt{3}}$
$→\left\{{{\frac{1}{3}=\frac{1}{m^2}}\atop{\frac{1}{3}=\frac{m}{3\sqrt{3}}}}\right.$
$→\left\{{{m^2=3}\atop{m=\sqrt{3}}}\right.$
$→m=\sqrt{3}$