10．B 点拨：过P作PF∥BC交AC于F.∵△ABC为等边三角形，∴易得△APF也是等边三角形，∴AP＝PF.∵AP＝CQ，∴PF＝CQ.又PF∥CQ，∴∠DPF＝∠DQC，∠DFP＝∠DCQ，∴△PFD≌△QCD.∴DF11111＝DC.∵PE⊥AF，且PF＝PA，∴AE＝EF.∴DE＝DF＋EF＝CF＋AF＝AC＝×1＝.
22222
1
二、11.；－32a10b5 12.(－2，－3)
813．(a＋b)(a－3b) 14.8 15.55°16．100°
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17．10 点拨：利用正多边形的性质可得点F关于直线AD的对称点为点B，连接BE交AD于点P′，连接FP′，那么有P′B＝P′F.所以P′E＋P′F＝P′E＋P′B＝BE.当点P与点P′重合时，PE＋PF的值最小，最小值为BE的长．易知△AP′B和△EP′F均为等边三角形，所以P′B＝P′E＝5，可得BE＝10.所以PE＋PF的最小值为10.
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－
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－?＋22．解：(1)原式＝4－a2＋a2－2ab＋3a5b÷a8b4＝4－2ab＋3a3b3.当ab＝－时，原式＝4－2×??2?213
－?＝4＋1－3＝5－24＝－19.3×??2??1??2?－3
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(2)原式＝a[(n－1)2－2(n－1)＋1]＝a(n－1－1)2＝a(n－2)2.
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23．解：(1)方程两边乘(x－3)，得1－2(x－3)＝－3x，解得x＝－7.检验：当x＝－7时，x－3≠0，∴原分式方程的解为x＝－7.
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(2)方程两边同乘2x(x＋1)得3(x＋1)＝4x，解得x＝3.检验：当x＝3时，x≠0，x＋1≠0，∴原分式方程的解为x＝3.
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ay use the contents or services of this article fo24．解：(1)A(－3，3)，B(－5，1)，C(－1，0)．
(2)图略，关于y轴对称的两个点横坐标互为相反数，纵坐标相等(两点连线被y轴垂直平分)． 111
(3)S△ABC＝3×4－×2×3－×2×2－×4×1＝5.
2221
25．解：BE＝DF.证明如下．
2
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ay use the contents or services of this article for如图，过点D作DH∥AC，交BE的延长线于点H，交AB于点G.
(第25题)
∵DH∥AC，∴∠BDH＝∠C.1
∵∠EDB＝∠C，
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∴∠EDB＝∠BDH.
2∴∠EDB＝∠EDH. 在△EDB与△EDH中， ∠EDB＝∠EDH，??
?ED＝ED，?，?∠BED＝∠HED＝90°∴△EDB≌△EDH. 1∴BE＝HE，即BE＝BH.
2∵AB＝AC，∠BAC＝90°，∴∠ABC＝∠C＝45°. 又∵DH∥AC，
∴∠BGD＝90°，∠BDG＝45°. ∴BG＝DG，∠BGH＝∠DGB＝90°.又∵BE⊥DE，∠BFE＝∠DFG， ∴∠GBH＝∠GDF. ∴△GBH≌△GDF.∴BH＝DF. 1∴BE＝DF. 2
1
点拨：通过添加辅助线，易得△EDB≌△EDH，也就是通过构造轴对称图形得到BE＝EH＝BH，此
2为解答本题的突破口．
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7 500116 000
＝×，x2x＋10
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解得x＝150，经检验，x＝150是原方程的解，且符合题意． 答：第二批鲜花每盒的进价是150元．
???a＋b－3＝0，?a＝2，
27．(1)证明：∵|a＋b－3|＋(a－2b)＝0，∴?解得?∴A(1，3)，B(2，0)．作AE⊥OB
?a－2b＝0，?b＝1.??
2
于点E，∵A(1，3)，B(2，0)，∴OE＝1，BE＝2－1＝1，在△AEO与△AEB中，
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AE＝AE，??
，∵?∠AEO＝∠AEB＝90°
??OE＝BE，
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∴△AEO≌△AEB，∴OA＝AB. (2)证明：∵∠CAD＝∠OAB， ∴∠CAD＋∠BAC＝∠OAB＋∠BAC，
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OA＝AB，??
即∠OAC＝∠BAD.在△AOC与△ABD中，∵?∠OAC＝∠BAD，
??AC＝AD，∴△AOC≌△ABD.
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(3)解：点P在y轴上的位置不发生改变．理由：设∠AOB＝α.∵OA＝AB，∴∠AOB＝∠ABO＝α.由(2)知，
△AOC≌△ABD，∴∠ABD＝∠AOB＝α.∵OB＝2，∠OBP＝180°－∠ABO－∠ABD＝180°－2α为定值，∠POB＝90°，易知△POB形状、大小确定，∴OP长度不变，∴点P在y轴上的位置不发生改变． 路上，每个人的相遇、相识，都是一份缘，我们都是相互之间不可或缺的伴。 版权申明 本文部分内容，包括文字、图片、以及设计等在网上搜集整理.版权为个人所有 This article includes some parts, including text, pictures, and design. Copyright is personal ownership. 用户可将本文地内容或服务用于个人学习、研究或欣赏，以及其他非商业性或非盈利性用途，但同时应遵守著作权法及其他相关法律地规定，不得侵犯本网站及相关权利人地合法权利.除此以外，将本文任何内容或服务用于其他用途时，须征得本人及相关权利人地书面许可，并支付报酬. Users may use the contents or services of this article for personal study, research or appreciation, and other non-commercial or non-profit purposes, but at the same time, they shall abide by the provisions of copyright law and other relevant laws, and shall not infringe upon the legitimate rights of this website and its relevant obligees. In addition, when any content or service of this article is used for other purposes, written permission and remuneration shall be obtained from the person concerned and the relevant obligee. 转载或引用本文内容必须是以新闻性或资料性公共免费信息为使用目地地合理、善意引用，不得对本文内容原意进行曲解、修改，并自负版权等法律责任.Reproduction or quotation of the content of this article must be reasonable and good-faith citation for the use of news or informative public free information. It shall not misinterpret or modify the original intention of the content of this article, and shall bear legal liability such as copyright.
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