若直线y=kx+3与直线y=(1/k)x-5的交点第一象限,则k的取值范围是
由kx+3=(1/k)x-5,得(k-1/k)x=[(k²-1)/k]x=[(k+1)(k-1)/k]x=-8,即有x=-8k/(k+1)(k-1)>0,
故得k