In order to compare the size with 1,1(a-2) in the third bracket is rewritten as-1/(2-a).

① when a & ltAt 1,1(2-a) <1,where the solution is x; 1; X ∈ (-∞， 1/(2-a))∨( 1，+∞)；

② When a= 1, the formula (1) becomes: (x- 1)? ≧0, and the solution at this time is: x ∈ r;

③ When 1

(b) In B)2005, when a>2 o'clock direction is A-2 > 0, if both sides are divided by (a-2), and the direction of unequal signs remains unchanged, we get: (x-1) [x+1(a-2)] ≦ 0;

The solution at this time is:-1/(a-2) ≦ x ≦1;