smc

Multiplying both sides by $(x-1)(m-1)m$ gives us $$xm(x-1)-m(m+1)=x(x-1)(m-1)$$ $$x^{2}m-xm-m^2-m=x^{2}m-xm-x^2+x$$ $$-m^2-m=-x^2+x$$ $$x^2-x-(m^2+m)=0$$ The roots of this quadratic are equal if and only if its discriminant ($b^2-4ac$) evaluates to 0. This means $$(-1{)}^2-4(-m^2-m)=0$$ $$4m^2+4m+1=0$$ $$(2m+1{)}^2=0$$ $$m=-\frac{1}{2}\Rightarrow \boxed{m=-\frac{1}{2}}$$