61. 解：原式······························································· 2分

·················································· 2分

原式　

60. 解：.　 ………………………4分

(其中通分1分，除法变乘法1分，分子分母分解因式1分，化简1分)

在-3 < p < 3中的整数p是-2，-1，0，1，2， ……………………………………………………5分

根据题意，这里p仅能取-1，此时原式 = .…………………………………………………6分

(若取p = -2，0，1，2，代入求值，本步骤不得分；直接代-1计算正确给1分)

59. 解：原式=

　　　　　　　　 =………………2分

　　　　　　　　 =x－4 ………………3分

　　　　　　　　 当时， 原式=………………4分

58. 解：原式＝　　

=

=．　　　

当，时，

原式=

57. 解：原式··········································································· 4分

························································································································· 6分

当时，原式．································································································· 7分

56.

·············································································································· 4分

．······················································································································ 6分

当时，

．··············································································· 8分

55.

(2)原式=

==

54. 解法一：设乙同学的速度为米/秒，则甲同学的速度为米/秒，

根据题意，得，

解得．

经检验，是方程的解，且符合题意．

甲同学所用的时间为：(秒)，

乙同学所用的时间为：(秒)．

，乙同学获胜．

解法二：设甲同学所用的时间为秒，乙同学所用的时间为秒，

根据题意，得

解得

经检验，，是方程组的解，且符合题意．　　　　　　　　　　　　　　

，乙同学获胜．

53. 解: (以下给出一种方案，其他方案略)

　 = =

52. 解：设第一天捐款x人，则第二天捐款x+50人，由题意列方程 　　　　　　　　　　　　　　　　　

　　　　 = ．　

　　　 化简得，4x+200=5x．　 　解得 x =200．

检验：当x =200时，x(x+50)≠0，　

　　　 ∴ x =200是原方程的解．　　　　

　　 两天捐款人数x+(x+50)=450．

人均捐款=24．　

答：两天共参加捐款的有450人；人均捐款24元