Question

设f(x)=|x-3|+|x-4|，解不等式f(x)<=2

Answer 1

X大于4，f（x）=2x-7<=2则x=4；x在3、4之间。则x=3；x小于3，则x=3

Answer 2

解析：当 x<3时，f(x) =3-x+4-x = 7-2x，故 f(x) <=2推出5/2<= x<3；
当 3 <= x <4时，f(x) = x-3+4-x = 1，推出 3<=x <4；
当 x>=4时，f(x) = x-3 + x-4 = 2x-7 <=2，推出 4<=x <=9/2 于是解集为 {x| 5/2<=x<=9/2}

Answer 3

讨论啊，分成三个区间，当x<=3时， 3-x+4-x<=2, 7-2x<=2, 5<=2x, 5/2<= x<=3当3<x<4时， x-2+4-x<=2, 2<=2, 恒成立，2<x<4当x>=4时， x-3+x-4<=2, 2x-7<=2, 4<=2x<=9综合上述：（2，9】希望得到您100%的好评，谢谢！

Answer 4

[2.5 , 4.5]