8. 解：⑴设所围矩形ABCD的长AB为x米，则宽AD为米．

依题意，得　

即，　

解此方程，得　　 　

∵墙的长度不超过45m，∴不合题意，应舍去．　

当时，

所以，当所围矩形的长为30m、宽为25m时，能使矩形的面积为750m²．

⑵不能．因为由得

又∵＝(－80)²－4×1×1620=－80＜0，

∴上述方程没有实数根．

因此，不能使所围矩形场地的面积为810m²

7. 解：由题意，△=(-4)²-4(m-)=0

即16-4m+2=0，m=．

当m=时，方程有两个相等的实数根x₁=x₂=2．

6. 解：

5. 解：(1)设A市投资“改水工程”年平均增长率是x，则

解之，得x＝0.4或x＝－2.4(不合题意，舍去)

所以，A市三年共投资“改水工程”2616万元.

4. 解法一：因为，所以．·················· 3分

即．所以，原方程的根为，．························· 6分

解法二：配方，得．··················································································· 2分

直接开平方，得．····················································································· 4分

所以，原方程的根为，．　 6分

3. 解：　　　 ………………1分

　　 　　　　　………………2分

　　　　　 ………………3分

∴x－1＝或x－1＝－ ………………4分

∴＝1+，＝1－　 ………………6分

2. x₁＝2　x₂＝　

1.

解：·································································································· 3分

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

，······································································································· 6分

1. ；2. -1；3. ；4. 10;5. ;6. ;7. 6或10或12;　 8.　 0；　 9.　 ，；　 10. 7，3；11. ，；　 12. ； 13.　 4；　 14.　 ；15. －4；16.　 4；　 17. 2；18. 　　19. 10%；　 20.　 +40－75=0 ； 21.、

1.A 2.D 3.A 4.B 5.C 6.C 7.B 8.D 9.B 10.C 11.A 12.A 13.A 14.A 15.D 16.D 17.D 18.D 19.B 20.A 21.C 22.A 23.B 24.B 25.B 26.D 27.C 28.B 29.A 30.B 31.C 32.A