搜题
问题   更新时间2023/4/3 12:59:00

解不等式x2-4x+3/x2-7x+10>0

解x2-4x+3/x2-7x+10=(x-1)(x-3)/(x-2)(x-5=(x-1)(x-3)(x-2)(x-5)/(x-2)2(x-5)2>0 即(x-1)(x-3)(x-2)(x-5)>0, 即x<1,或2<x<3或x>5时(x-1)(x-3)(x-2)(x-5)>0 故x2-4x+3/x2-7x+10>0的解集为{x|(﹣∞,1)∪(2,3)∪(5,﹢∞)}
王老师:19139051760(拨打)