10. 解答： 又，

　当时，的值最小

此时，即最小值为．

9. 解法一：设矩形温室的宽为xm,则长为2xm.根据题意，得

(x-2)·(2x-4)=288. …………………………………………………4分

解这个方程，得

x₁=-10(不合题意，舍去)，x₂=14……………………………………6分

所以x=14,2x=2×14=28.

答：当矩形温室的长为28m，宽为14m时，蔬菜种植区域的面积是288m².

……………………………………………………………………………7分

解法二：设矩形温室的长为xm，则宽为xm,根据题意，得

(x-2)·(x-4)=288. ………………………………………………4分

解这个方程，得

x₁=-20(不合题意，舍去)，x₂=28. ……………………………………6分

所以x=28, x=×28=14.

答：当矩形温室的长为28m，宽为14m时，蔬菜种植区域的面积是288m².

……………………………………………………………………………7分

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