(1)由题意知,k≠0且△=b2-4ac>0∴b2-4ac=[-2(k+1)]2-4k(k-1)>0,即4k2+8k+4-4k2+4k>0,∴12k>-4解得:k>- 1 3 且k≠0(2)不存在.∵x1+x2= 2(k+1) k ,x1?x2= k?1 k ,又有 1 x1 + 1 x2 = x1+x2 x1x2 =1,可求得k=-3,而-3<- 1 3 ∴满足条件的k值不存在.