Question

2．若a=$\sqrt{5}$+3，则2a³-11a²+2a+1=-3．

Answer 1

分析 把2a³-11a²+2a+1变形为2a（a-3）²+a²-16a+1，代入a=$\sqrt{5}$+3，得出a²-6a+1，进一步变形为（a-3）²-8，代入a=$\sqrt{5}$+3即可求得代数式的值．

解答 解：2a³-11a²+2a+1
=2a³-12a²+18a+a²-16a+1
=2a（a-3）²+a²-16a+1
=2a×5+a²-16a+1
=a²-6a+1
=a²-6a+9-8
=（a-3）²-8
=5-8
=-3．
故答案为-3．

点评 本题考查了二次根式的化简求值，把代数式进行两次变形是解题的关键．