令f(x)=x^2+2mx+2m-2,由已知有：（1）f(-2)＞0,即4-4m+2m-2＞0,得m＜1(2) f(2)＞0,即4+4m+2m-2＞0,得m＞-1/3(3) △=4m^2-4(2m-2)＞0,恒成立所以-1/3＜m＜1注：函数类题目关键要根据根的条件画出草图（此题二次系数＞...