13.、解：

　　　　 =············································································ 3分

或 ····························································································· 5分

当时，分式无意义．

原式的值为2．······························································································ 6分

12. ．原式=[–]×······················································ 3分

=×–×

=–·································································································· 4分

=–

=··········································································································· 5分

当x=1时，

原式=····································································································· 6分

= 1················································································································· 7分

说明：以上步骤可合理省略

11. 解：原式=

　　　　　　　　 =………………2分

　　　　　　　　 =x－4 ………………3分

　　　　　　　　 当时， 原式=………………4分

10.

解:原式=…=　　　

由得，

代入原式=2　

9. 解:原式=　　　

8. 解：

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

········································································································· 3分

．···················································································································· 4分

当时，原式．······································································ 6分

7. 解：(以下给出三种选择方案，其他方案从略)　

　 　解答一：Y + Z =(3a²+3ab)+ (a²+ab)

　　　　　　　 =4a²+4ab　　　

　　　　　　　 =4a(a+b)．　　　

　 　解答二： X- Z = (2a²+3ab+b²)-(a²+ab)

　　　　　　 　=a²+2ab+b²=(a+b)²．　　　

解答三：Y- X=(3a²+3ab)- (2a²+3ab+b²)

=a²- b² =(a+b)(a-b)．　

6. 解：原式==

5.

解：原式=4a²+4a+1-4a-2+3……………………………………………………3分

　　　　 =4a²+2.…………………………………………………………………4分

当a=时，4a²+2=4×()²+2=10. …………………………………6分

4.

解：解：原式

　　 将代入上式得

　　 原式