设：a>1,-1<a<1,a<-1;
|ax-1|>=x等价于ax-1>=x或ax-1<=-x;即（a-1）x>=1或(a+1)x<=1
(1)a>1时,a-1>0,a+1>0，所以x>=1/(a-1),或x<=1/(a+1),由于a-1<a+1,该解不存在，即a>1时无解；
(2)-1<a<1时，a-1<0，a+1>0,所以有 x<=1/(a-1),或x<=1/(a+1),由于a-1<a+1，所以解得想x<=1/(a+1);
(3)a<-1时,a-1<0，a+1<0,所以有 x<=1/(a-1),或x>=1/(a+1),，得1/(a+1)<x<1/(a-1)
|ax-1|>=x等价于ax-1>=x或ax-1<=-x;即（a-1）x>=1或(a+1)x<=1
(1)a>1时,a-1>0,a+1>0，所以x>=1/(a-1),或x<=1/(a+1),由于a-1<a+1,该解不存在，即a>1时无解；
(2)-1<a<1时，a-1<0，a+1>0,所以有 x<=1/(a-1),或x<=1/(a+1),由于a-1<a+1，所以解得想x<=1/(a+1);
(3)a<-1时,a-1<0，a+1<0,所以有 x<=1/(a-1),或x>=1/(a+1),，得1/(a+1)<x<1/(a-1)
