(x²+kx-2)/(x²-x+1)＜2
(x²+kx-2)/(x²-x+1)-2＜0
(x²+kx-2)/(x²-x+1)-2(x²-x+1)/(x²-x+1)＜0
[(x²+kx-2)-2(x²-x+1)]/(x²-x+1)＜0
[-x²+(k+2)x-4]/(x²-x+1)＜0
因为分母x²-x+1=(x-1/2)²+3/4＞0恒成立
所以不等式等价于-x²+(k+2)x-4＜0
即x²-(k+2)x+4＞0恒成立
所以判别式Δ=(k+2)²-16＜0
所以(k+2)²＜16
-4＜k+2＜4
-6＜k＜2
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