由于x2+2＞1，∴不等式log(x2+2)(3x2−2x−4)＞log(x2+2)(x2−3x+2)可化为：3x2−2x−4＞0x2−3x+2＞03x2−2x−4＞x2−3x+2⇒x2−3x+2＞03x2−2x−4＞x2−3x+2⇒x2−3x+2＞02x2+x−6＞0解得：x＞2或x＜-2故原不等式...