20 0 1年全国高中数学竞赛第一试第 11题为 :函数 y =x + x2 - 3 x+ 2的值域为.下面提供五种解法 ,以飨读者 .解法 1　移项得 y- x=x2 - 3 x+ 2 ,上式等价于 (y- x) 2 =x2 - 3 x+ 2 ,y- x≥ 0 .12由 1得 x=y2 - 22 y- 3 ,代入 2得 y- y2 - 22 y- 3≥ 0 ,即 (y- 1) (y- 2 )2 y- The first question of the first trial of the National High School Mathematics Competition in 2001 was: The range of the function y =x + x2 - 3 x + 2 is. The following five solutions are provided to readers. Solution 1 y-x= X2 - 3 x+ 2 , the above equation is equivalent to (y- x) 2 =x2 - 3 x+ 2 ,y- x ≥ 0 .12 from 1 to x=y2 - 22 y- 3 , substituting 2 for y- y2 - 22 y- 3≥ 0, that is (y- 1) (y- 2 )2 y-