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8. 解：⑴設(shè)所圍矩形ABCD的長(zhǎng)AB為x米，則寬AD為米．

依題意，得　

即，　

解此方程，得　　 　

∵墻的長(zhǎng)度不超過(guò)45m，∴不合題意，應(yīng)舍去．　

當(dāng)時(shí)，

所以，當(dāng)所圍矩形的長(zhǎng)為30m、寬為25m時(shí)，能使矩形的面積為750m²．

⑵不能．因?yàn)橛?i>得

又∵＝(－80)²－4×1×1620=－80＜0，

∴上述方程沒(méi)有實(shí)數(shù)根．

因此，不能使所圍矩形場(chǎng)地的面積為810m²

7. 解：由題意，△=(-4)²-4(m-)=0

即16-4m+2=0，m=．

當(dāng)m=時(shí)，方程有兩個(gè)相等的實(shí)數(shù)根x₁=x₂=2．

6. 解：

5. 解：(1)設(shè)A市投資“改水工程”年平均增長(zhǎng)率是x，則

解之，得x＝0.4或x＝－2.4(不合題意，舍去)

所以，A市三年共投資“改水工程”2616萬(wàn)元.

4. 解法一：因?yàn)?sub>，所以．·················· 3分

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

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

直接開(kāi)平方，得．····················································································· 4分

所以，原方程的根為，．　 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