当x＝2时，原式＝2×(2)2＋1＝5． ··················································· 7分 3x?2?x?1)?2x?2x?1 2、x?1(3x2?1x?2答案：原式?(?)?x?1x?1(x?1)2 ………………………………1分

?(x?2)(x?2)(x?1)2??x?1x?2 …………………………………………3分

??x2?x?2 …………………………………………4分

3、（2020昆山一模）计算：已知x2－4＝0，求代数式x(x＋1)2－x(x2＋x)－x－7的值． 答案：

4、(2020年北京市延庆县一诊考试)化简求值：当2x2?3x?1?0时 ，求

(x?2)2?x(x?5)?2x?8的值.

解：?x?2??x?x?5??2x?8 =x2?4x?4?x2?5x?2x?8 =2x2?3x?4 ∵2x2?3x?1=0

∴2x2?3x??1

∴原式=2x2?3x?4=?1?4??5

25、(2020年福州模拟卷) (2) 先化简，再求值：(x＋1)2＋x(x－2)，其中x＝2． (2)解：(x＋1)2＋x(x－2) ＝x2＋2x＋1＋x2－2x ······································································· 4分 ＝2x2＋1， ···················································································· 5分

当x＝2时，原式＝2×(2)2＋1＝5． ··················································· 7分