2018年广州市有关中考数学真题和详细解析.pdf

2018年广州市有关中考数学真题和详细解析（纯word 版）2011 年广州市初中毕业生学业考试一、选择题（每小题3 分，共 30 分）1.四个数-5，21，3中为无理数的是（）A.-5 B.C.21 D.32.已知ABCD 的周长为32，AB=4，则 BC=（）A.4 B.121 C.24 D.283.某车间 5 名工人日加工零件数分别为6，10，4，5，4，则这组数据的中位数是()A.4 B.5 C.6 D.104.将点 A（2，1）向左平移2 个单位长度得到点A，则点A的坐标是（）A.（0，1）B.（2，-1）C.（4，1）D.（2，3）5.下列函数中，当x0 时，y 值随 x 值增大而减小的是（）A.2xy B.1xy C.xy43 D.xy16.若 ac0b，则 abc 与 0 的大小关系是（）A.abc0 D.无法确定7.下面的计算正确的是（）A.2221243xxx B.1553xxx C.34xxx D.725)(xx8.如图所示，将矩形纸片先沿虚线AB 按箭头方向向右对折，接着对折后的纸片沿虚线CD向下对折，然后剪下一个小三角形，再将纸片打开，则打开后的展开图是（）9.当实数 x 的取值使得2x有意义时，函数y=4x+1 中 y 的取值范围是（）-7 B.y9 C.y9 D.y910.如图，AB切 O于点 B，OA=23，AB=3，弦 BC33 B.23 C.D.23二、填空题：（每小题3 分，共 18 分）的相反数是 _（12.已知=260，则的补角是 _度。13.方程231xx的解是 _14.如图，以点O为位似中心，将五边形ABCDE 放大后得到五边形EDCBA，已知 OA=10cm，AO=20cm，则五边形ABCDE的周长与五边形EDCBA的周长的比值是_15.已知三条不同的直线a、b、c 在同一平面内，下列四条命题：如果 a 其中真命题的是_。（填写所有真命题的序号）16.定义新运算“”，baba431，则)1(12=_。三、解答题（本大题共9 大题，满分102 分）17.（9 分）解不等式组01231xx18.（9 分）如图，AC是菱形 ABCD 的对角线，点E、F 分别在边AB、AD上，且 AE=AF。求证：ACE ACF19.（10 分）分解因式：8(x2-2y2)-x(7x+y)+xy20.（10 分）5 个棱长为1 的正方体组成如图的几何体。（1）该几何体的体积是_(立方单位)表面积是 _(平方单位)（2）画出该几何体的主视图和左视图。21.(12分)某商店 5月 1 日举行促销优惠活动，当天到该商店购买商品有两种方案，方案一：用 168 元购买会员卡成为会员后，凭会员卡购买商店内任何商品，一律按商品价格的8折优惠；方案二：若不购买会员卡，则购买商店内任何商品，一律按商品价格的折优惠。已知小敏 5 月 1 日前不是该商店的会员。（1）若小敏不购买会员卡，所购买商品的价格为120 元时，实际应支付多少元（2）请帮小敏算一算，所购买商品的价格在什么范围时，采用方案一更合算ADFEBC正面文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 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ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W122.（12 分）某中学九年级（3）班 50 名学生参加平均每周上网时间的调查，由调查结果绘制了频数分布直方图，根据图中信息回答下列问题：（1）求 a的值；（2）用列举法求以下事件的概率：从上网时间在610 小时的 5 名学生中随机选取2 人，其中至少有 1 人的上网时间在810 小时。23.（12 分）已知RtABC的斜边 AB在平面直角坐标系的x 轴上，点C(1,3)在反比例函数y=xk的图象上，且sin BAC=53。（1）求 k 的值和边AC的长；（2）求点 B的坐标。24.（14 分）已知关于x 的二次函数y=ax2+bx+c(a0)的图象经过点C(0,1)，且与 x 轴交于不同的两点A、B，点 A的坐标是（1，0）（1）求 c 的值；（2）求 a的取值范围；（3）该二次函数的图象与直线y=1 交于 C、D两点，设A、B、C、D四点构成的四边形的对角线相交于点P，记 PCD的面积为S1，PAB的面积为S2，当 0a1 时，求证：S1-S2为常数，并求出该常数。25.（14 分）如图 7，O中 AB是直径，C是 O上一点，ABC=450，等腰直角三角形DCE中 DCE是直角，点D在线段 AC上。（1）证明：B、C、E三点共线；文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 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HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1（2）若 M是线段 BE的中点，N是线段 AD的中点，证明：MN=2OM；（3）将 DCE绕点 C逆时针旋转（00900）后，记为 D1CE1（图 8），若 M1是线段 BE1的中点，N1是线段 AD1的中点，M1N1=2OM1是否成立若是，请证明：若不是，说明理由。2011 年广州市中考数学试题答案一、选择题文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W11、D 2、B 3、B 4、A 5、D 6、C 7、C 8、D 9、B 10、A二、填空题11、9；12、154；13、1x；14、12；15、；16、8。三、解答题17、解：解不等式，得4x解不等式，得12x 不等式组的解集为142x18、证明：AC 是菱形 ABCD 的对角线 CAE=CAF在 ACE和 ACF中 AE=AF，CAE=CAF，AC=AC ACE ACF19、解：22827xyxxyxy2228167xyxxyxy2216xy44xyxy20、解：(1)5，20；（2）21、解：(1)实际应支付：120 114（元）(2)设所购商品的价格为x 元，依题意得 168解得 x 1120 当所购商品的价格高于1120 元时，选方案一更合算。22、解：(1)506253214a (2)将上网时间在68 小时的 3人记为 A、B、C，上网时间在810 小时的 2 人记为D、E，从中选取2 人的所有情况为（A、B）、（A、C）、（A、D）、（A、E）、（B、C）、（B、D）、（B、E）、（C、D）、（C、E）、（D、E）共 10 种等可能的结果，其中至少有一人上网时间在在810 小时的有（A、D）、（A、E）、（B、D）、（B、E）、（C、D）、（C、E）、（D、E）这 7种，所以至少有一人上网时间在在810 小时的概率为0.7。左视图主视图文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W123、解：(1)点 A（1,3）在反比例函数kyx的图像上1 33kxy作 CD AB于点 D，所以 CD 3在 RtACD中，sin BAC=CDAC，335AC，解得 AC=5(2)在 RtACD中，2222534ADACCD cosBAC=45ADAC如图 1，在在 Rt ACD 中，cos BAC=ACAB，2554cos45ACABBAC4 13AOADOD2513344OBABOA 点 B的坐标为13,04如图 2，415AOADOD255544OBABOA 点 B的坐标为5,0424、解：(1)将点 C（0，1）代入2yaxbxc得1c（2）由(1)知21yaxbx，将点 A（1，0）代入得10ab，1ba 二次函数为211yaxax二次函数为211yaxax的图像与x 轴交于不同的两点0，而222214214211aaaaaaaaa的取值范围是0a且1axy图1BACODxy图2BACOD文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1 (3)证明：01a 对称轴为11122aaxaa11212aaABaa把1y代入211yaxax得210axax，解得1210,axxa1aCDa12PCDPABACDCABSSSSSS1122CDOCABOC111111222aaa112SS为常数，这个常数为1。25、（1）证明：AB 是 O的直径 ACB=90 DCE=90 ACB DCE=180 B、C、E三点共线。(2)证明：连接ON、AE、BD，延长 BD交 AE于点 FABC=45，ACB=90 BC=AC，又 ACB=DCE=90，DC=EC BCD ACE BD=AE，DBC=CAE DBC AEC=CAE AEC=90 BFAE AO=OB，AN=ND ON=12BD，ON BD AO=OB，EM=MB OM=12AE，OM AE OM=ON，OM ON OMN=45，又 cos OMN=OMMN2MNOM(3)1112M NOM成立，证明同（2）。xyPDBCOAFNMDOBCAEFN1M1DOBCAE文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 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ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档编码:CR7G3I8R3M7 HU2G7I8B7X6 ZI4A5O6Z3W1文档