已知关于x的一元二次方程kx2-2（k+1）x+k-1=0有两个不相等的实数根x1，x2．（1）求k的取值范围；（2）是
推荐回答（1个）
（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＞-且k≠0
（2）不存在．
∵x1+x2=，x1?x2=，
又有+==1，
可求得k=-3，而-3＜-
∴满足条件的k值不存在．
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