函数f(x)=ax/(x²-1).a＞0.因m∈(0,1),∴-1＜-m＜m＜1.∴在区间[-m,m]上,函数f(x)有意义.求导得f'(x)=-a(x²+1)/(x²-1)²＜0.∴在[-m,m]上,函数f(x)递减,由题设应有f(-m)=m,且f(m)=-m.===>am/(m²-1)=-m.===>a=1-m².∵m∈(0,1).∴0＜a＜1.