∵|2x+1|＜1∴-1＜2x+1＜1∴-2＜2x＜0∴-1＜x＜0下式可化为：x²-2x-5＜-2，即：x²-2x-3＞0x -3x 1由十字相乘法，可得：（x-3）（x+1）＞0解得：x＞3或x＜-1∵-1＜x＜0且x＞3或x＜-1∴x无解故，不等式组的解集为空集。答题不易，望采纳~~~~~
4x+6﹥1一x3(x一1)≤x+5
