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2009年辽宁省朝阳市中考数学试题及答案(word版)



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2009 年辽宁省朝阳市初中升学考试 数 学 试 卷
一、选择题(下列各题的备选答案中,只有一个是正确的,请将正确答案的序号填入下面表 选择题(下列各题的备选答案中,只有一个是正确的, 格内, 格内,每小题 3 分,共 24 分) 1.2 的倒数的相反数是( A. )

1 C.2 D. ?2 2 2.如图,已知 AB ∥ CD ,若 ∠A = 20° ∠E = 35° , ,则 ∠ C
B. ? 等于( A.20° ) B.35° C.45° D.55° ) A F C

1 2

E B D (第 2 题图)
3

3. 某市水质检测部门 2008 年全年共监测水量达 28909.6 万吨. 将 数字 28909.6 用科学记数法 (保留两位有效数字) 表示为 ( A. 2.8 × 10 B. 2.9 × 10 4.下列运算中,不正确的是( )
4 4

C. 2.9 ×10

5

D. 2.9 ×10
3

A. a + a = 2a
3 3

3

B. a · = a a
2 3

5

C. ( ?a 3 ) 2 = a 9 )

D. 2a ÷ a = 2a
2

5.如图是某体育馆内的颁奖台,其左视图是(

A. . 6.下列事件中,属于不确定事件的有( ④小明长大后成为一名宇航员 A.①②③ B.①③④ ) C.②③④ )

B.

C.

D.

① 太阳从西边升起;② 任意摸一张体育彩票会中奖;③ 掷一枚硬币,有国徽的一面朝下;
D.①②④

7.下列说法中,正确的是( A.如果

a+b c+d a c = ,那么 = b d b d

B. 9 的算术平方根等于 3 D.方程 x + x ? 2 = 0 的根是 x1 = ?1,x 2 = 2
2

C.当 x < 1 时, x ? 1 有意义 8.下列命题中,不正确的是( ) A. n 边形的内角和等于 ( n ? 2) 180° ·

B.边长分别为 3, 5, 4, 的三角形是直角三角形 C.垂直于弦的直径平分弦所对的两条弧

温度(℃) 30 28 26 24 22 20 18 版权所有@中国教育考试资源网 日期 16 1 2 3 4 5 6 7 8 9 10

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D.两圆相切时,圆心距等于两圆半径之和 二、填空题(每小题 3 分,共 24 分) 填空题(

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9.如图是某地 5 月上旬日平均气温统计图,这些气温数据的众数是__________,中位数是 __________,极差是__________. 10.如图, △ ABC 是等边三角形,点 D 是 BC 边上任意一点, A

DE ⊥ AB 于 点 E , DF ⊥ AC 于 点 F . 若 BC = 2 , 则 DE + DF = _____________. 11.如图是小明从学校到家里行进的路程 S (米)与时间 t (分) E 的函数图象.观察图象,从中得到如下信息: ① 学校离小明家 B 1000 米; ② 小明用了 20 分钟到家; ③ 小明前 10 分钟走了路
程的一半;④小明后 10 分钟比前 10 分钟走的快,其中正确 的有___________(填序号) . 12.如图, AC 是汽车挡风玻璃前的刮雨刷.如果 AO = 65cm , 1000

F D (第 10 题图) s(米) C

CO = 15cm ,当 AC 绕点 O 旋转 90°时,则刮雨刷 AC 扫
过的面积为____________cm2. ,一条对角线的长为 2 3 ,则另 13.已知菱形的一个内角为 60° 一条对角线的长为______________. 14.如图,正比例函数 y = 0 t(分) 10 20 (第 11 题图)

k (k ≠ 0) x 的图象在第一角限内交于点 A ,且 AO = 2 ,则 k = ____________.
3 x 与反比例函数 y =

15.如图,路灯距离地面 8 米,身高 1.6 米的小明站在距离灯的底部(点 O )20 米的 A 处, 则小明的影长为___________米. y A′ A C′ C A O (第 12 题图) (第 14 题图) O x O B A (第 15 题图) M

16.下列是有规律排列的一列数: 1, , ,, ……其中从左至右第 100 个数是__________. (每小题 三、 每小题 8 分,共 16 分) ( 17.先化简,再求值:

3 2 5 3 4 3 8 5

x + 1 ? 1 + x2 ? ÷? x ? ? ,其中 x = 2 + 1 . x 2x ? ?
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, 18.在 10 × 10 的网格纸上建立平面直角坐标系如图所示,在 Rt△ ABO 中, ∠OAB = 90° 4) 且点 B 的坐标为 (3, . (1)画出 △OAB 向左平移 3 个单位后的 △O1 A1 B1 ,写出点 B1 的坐标; (2) 画出 △OAB 绕点 O 顺时针旋转 90° 后的 △OA2 B2 , 并求点 B 旋转到点 B2 时, B 经 点 过的路线长(结果保留 π ) y B

O

A x

(第 18 题图)

(每小题 四、 每小题 10 分,共 20 分) ( 19.袋中装有除数字不同其它都相同的六个小球,球上分别标有数字 1,2,3,4,5,6. (1)从袋中摸出一个小球,求小球上数字小于 3 的概率; (2)将标有 1,2,3 数字的小球取出放入另外一个袋中,分别从两袋中各摸出一个小球, 求数字之和为偶数的概率. (要求用列表法或画树状图求解)

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⊙ 点 点 20. 如图, O 是 Rt△ ABC 的外接圆, O 在 AB 上,BD ⊥ AB , B 是垂足,OD ∥ AC ,
连接 CD . 求证: CD 是 ⊙O 的切线. C D

A

O

B

(第 20 题图)

(每小题 五、 每小题 10 分,共 20 分) ( 21. 在改革开放 30 年纪念活动中, 某校学生会就同学们对我国改革开放 30 年所取得的辉煌成 就的了解程度进行了随机抽样调查,并将调查结果绘制成如图所示的统计图的一部分. 人数/人

了解很少 不了解 10% 10%很了解 基本了解 30%

30 25 20 15 10 5

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不了解 了解很少 基本了解 很了解 了解程度

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根据统计图中的信息,解答下列问题: (1)本次抽样调查的样本容量是___________.调查中“了解很少”的学生占_________%; (2)补全条形统计图; (3)若全校共有学生 1300 人,那么该校约有多少名学生“很了解”我国改革开放 30 年来取 得的辉煌成就? (4)通过以上数据分析,请你从爱国教育的角度提出自己的观点和建议.

22.海峡两岸实现“三通”后,某水果销售公司从台湾采购苹果的成本大幅下降.请你根据两 位经理的对话,计算出该公司在实现“三通”前到台湾采购苹果的成本价格.

“三通” 前买台湾苹果 的成本价格是今年的 2倍

同样用 10 万元采购台湾 苹果,今年却比“三通” 前多购买了 2 万公斤

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(每小题 六、 每小题 10 分,共 20 分) ( 方向航行,航行一段时间到达小岛 B 处后,又沿 23.一艘小船从码头 A 出发,沿北偏东 53° 着北偏西 22° 方向航行了 10 海里到达 C 处,这时从码头测得小船在码头北偏东 23° 的方 向上,求此时小船与码头之间的距离( 2 ≈ 1.4,3 ≈ 1.7 ,结果保留整数) .

C 北 22° 北 23° 53° A (第 23 题图) B

24.某学校计划租用 6 辆客车送一批师生参加一年一度的哈尔滨冰雕节,感受冰雕艺术的魅 力.现有甲、乙两种客车,它们的载客量和租金如下表.设租用甲种客车 x 辆,租车总费 用为 y 元. 甲种客车 载客量(人/辆) 租金(元/辆) 45 280 乙种客车 30 200

(1)求出 y (元)与 x (辆)之间的函数关系式,指出自变量的取值范围; (2)若该校共有 240 名师生前往参加,领队老师从学校预支租车费用 1650 元,试问预支的
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租车费用是否可以结余?若有结余,最多可结余多少元?

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D(G)

C

A(E) F D

B 图① G C

A y (本题 七、 本题 12 分) ( 在梯形 ABCD 中, ∥ AB , ABC = 90° CD ∠ , 25. 如图 ① , D

F E 图② G

B C

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图③

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∠DAB = 60° AD = 2 , CD = 4 .另有一直角三角形 EFG , ∠EFG = 90° G 与点 D , ,点 重合,点 E 与点 A 重合,点 F 在 AB 上,让 △EFG 的边 EF 在 AB 上,点 G 在 DC 上,以 每秒 1 个单位的速度沿着 AB 方向向右运动,如图 ② ,点 F 与点 B 重合时停止运动,设运动 时间为 t 秒. (1)在上述运动过程中,请分别写出当四边形 FBCG 为正方形和四边形 AEGD 为平行四 边形时对应时刻 t 的值或范围; (2)以点 A 为原点,以 AB 所在直线为 x 轴,过点 A 垂直于 AB 的直线为 y 轴,建立如图 ③ 所示的坐标系.求过 A,D,C 三点的抛物线的解析式; (3)探究:延长 EG 交(2)中的抛物线于点 Q ,是否存在这样的时刻 t 使得 △ ABQ 的面 积与梯形 ABCD 的面积相等?若存在,求出 t 的值;若不存在,请说明理由.

y A

B O A′ x

B′图① (本题 八、 本题 14 分) ( 26.如图 ① ,点 A′ , B′ 的坐标分别为(2,0)和(0, ?4 ) , 将 △ A′B′O 绕点 O 按逆时针方向旋转 90° 后得 △ ABO ,点 y A D O E C 图② B x

A′ 的对应点是点 A ,点 B′ 的对应点是点 B .
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(1)写出 A , B 两点的坐标,并求出直线 AB 的解析式;

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(2)将 △ ABO 沿着垂直于 x 轴的线段 CD 折叠, C 在 x 轴上,点 D 在 AB 上,点 D 不 (点 如图 ② , 使点 B 落在 x 轴上, B 的对应点为点 E . 点 设点 C 的坐标为 x,) ( 0 , 与 A ,B 重合)

△CDE 与 △ ABO 重叠部分的面积为 S . i)试求出 S 与 x 之间的函数关系式(包括自变量 x 的取值范围) ; ii)当 x 为何值时, S 的面积最大?最大值是多少? iii)是否存在这样的点 C ,使得 △ ADE 为直角三角形?若存在,直接写出点 C 的坐标;
若不存在,请说明理由.

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参考答案及评分标准
一、选择题(每小题 3 分,共 24 分) 选择题( 题号 答案 1 B 2 D 3 B 4 C 5 D 6 C

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7 A

8 D

二、填空题(每小题 3 分,共 24 分) 填空题( 9.26,26,4 13.2 或 6 10. 3 14. 3 11.①②④ 15.5 16. 12.1 000 π

101 2 3 4 (原一列数可化为 、 、 、 200 2 4 6

5 、……) 8
17. (本题满分 8 分) 解:原式=

x + 1 2 x2 ? 1 ? x2 ÷ ······················································································ (2 分) x 2x

=

x +1 2x ······································································································ (4 分) x ( x + 1)( x ? 1)
2 . ·························································································································· (6 分) x ?1

=

将x=

2 + 1 代入上式得原式=

2 ( 2) 2 = = 2 . ······································· (8 分) 2 + 1 ?1 2 y
B1 B

18. (本题满分 8 分) 解: (1)画图 ························································· (1 分)

B1 (0, ·································································· (3 分) 4)
(2)画图 ······························································· (5 分) O1 A1 O A2

A x B2

Q OB = 32 + 42 = 5 ············································ (6 分)
∴ 点 B 旋转到点 B2 时,经过的路线长为
19. (本题满分 10 分) 解: (1)小于 3 的概率 P =

2× 5× π 5 = π. 4 2

(第 18 题图)

········································································································································· (8 分)

2 1 = ················································································· (4 分) 6 3
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(2)列表如下 1 4 5 6 5 6 7 2 6 7 8 3 7 8 9 1 2 3 4 5 6 4 5 6 4 5 6 5 和: 6 7 6 7 8 7 8 9 树状图如下 开始

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······························· (8 分)

从表或树状图中可以看出其和共有 9 种等可能结果, 其中是偶数的有 4 种结果, 所以和为偶数 的概率 P =

4 ················································································································ (10 分) 9

20. (本题满分 10 分) 证明:连接 CO ··············································································································· (1 分)

Q OD ∥ AC. COD = ∠ACO,∠CAO = ∠DOB ················································ (3 分) ∴∠ Q ∠ACO = ∠CAO ∴∠COD = ∠DOB ································································· (6 分) 又 OD = OD,OC = OB . ∴△COD ≌△BOD ······································································································ (8 分) ∴∠OCD = ∠OBD = 90° ∴ OC ⊥ CD ,即 CD 是 ⊙O 的切线 ··········································································· (10 分)
21. (本题满分 10 分) (1)50,50 ···················································································································· (4 分) (2)补图略 ···················································································································· (6 分) (3) 1300 × 10% = 130 人. (4) 由统计图可知, 不了解和了解很少的占 60%, 由此可以看出同学们对国情的关注不够. 建 议:加强国情教育、爱国教育等.本题答案不惟一,只要观点正确,建议合理即可. ····························· (10 分) 答:该校约有 130 名学生很了解我国改革开放 30 年来所取得的辉煌成就. ············· (8 分) 22. (本题满分 10 分) 解:设该公司今年到台湾采购苹果的成本价格为 x 元/公斤 ········································· (1 分) 根据题意列方程得

100000 100000 + = 20000 ·························································································· (5 分) x 2x 解得 x = 2.5 ···················································································································· (7 分) 经检验 x = 2.5 是原方程的根. ······················································································ (8 分) 当 x = 2.5 时, 2 x = 5 ···································································································· (9 分)
答:实现“三通”前该公司到台湾采购苹果的成本价格为 5 元/公斤. ···················· (10 分) 23. (本题满分 10 分) 解:由题意知: ∠BAC = 53° 23° = 30° ··································································· (1 分) ?

∠C = 23° 22° = 45° ··································································································· (3 分) +
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Q BC = 10

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过点 B 作 BD ⊥ AC ,垂足为 D ,则 CD = BD ·························································· (4 分)

∴ CD = BC cos 45° = 10 × ·

2 = 5 2 ≈ 7.0 ······························································ (6 分) 2

AD =

3 3 BC =5 2÷ =5 2× = 5 2 × 3 ≈ 5 × 1.4 × 1.7 ≈ 11.9 tan 30° 3 3

∴ AC = AD + CD = 11.9 + 7.0 = 18.9 ≈ 19 ································································ (9 分)
答:小船到码头的距离约为 19 海里 ············································································ (10 分) 24. (本题满分 10 分) (1) y = 280 x + (6 ? x ) × 200 = 80 x + 1200(0 ≤ x ≤ 6) ·········································· (4 分) (2)可以有结余,由题意知 ?

?80 x + 1200 ≤ 1650 ················································ (6 分) ?45 x + 30(6 ? x) ≥ 240

解不等式组得: 4 ≤ x ≤ 5

5 8

∴ 预支的租车费用可以有结余.···················································································· (8 分) Q x 取整数 ∴x取 4 或 5 Q k = 80 > 0 ∴ y 随 x 的增大而增大. ∴ 当 x = 4 时, y 的值最小. 其最小值 y = 4 × 80 + 1200 = 1520 元 ∴ 最多可结余 1650 ? 1520=130 元 ················································································ (10 分)
25. (本题满分 12 分) (1)当 t = 4 ? 3 时,四边形 FBCG 为正方形.······················································· (1 分) 当 0 < t ≤ 4 时,四边形 AEGD 为平行四边形.·························································· (2 分) (2)点 D 、 C 的坐标分别是( 1,3 )(5 ,3 ) ······················································ (4 分) ,

Q 抛物线经过原点 O (0,0)

∴ 设抛物线的解析式为 y = ax 2 + bx
将 D 、 C 两点坐标代入得

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?a + b = 3 ? ? ?25a + 5b = 3 ?

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? 3 ?a = ? ? 5 ··········································································· (6 分) 解得 ? 6 ?b = 3 ? 5 ? 3 2 6 x + 3 x ·································································· (7 分) 5 5

∴ 抛物线的解析式为 y = ?

(3)Q 点 Q 在抛物线上,∴ 点 Q ? x, ?

? ? ?

? 3 2 6 x + 3x ? ? 5 5 ?

过点 Q 作 QM ⊥ x 轴于点 M ,又 B (5, 0) 则 S△ ABQ =

1 5 3 2 6 AB QM = ·? · x + 3x 2 2 5 5

1 ? 3 x 2 + 6 3 x ········································································································ (8 分) 2 1 9 又 S四边形ABCD = (4 + 5) × 3 × = 3 ········································································ (9 分) 2 2 1 9 2 令 ? 3x + 6 3x = 3 2 2 Q EG 的延长线与抛物线交于 x 轴的上方 ∴? x 2 + 6 x = 9 解得 x = 3 ······················································································ (10 分)
= 当 x = 3 时, y = ?

3 6 9 ×9 + 3×3 = 3 5 5 5 MQ 9 9 = 3 ÷ 3 = . ············································ (11 分) tan 60° 5 5

Q ∠QEM = 60° EM = , ∴
∴t = 3 ? 9 6 = (秒) . 5 5

即存在这样的时刻 t ,当 t =

6 秒时, △ AQB 的面积与梯形 ABCD 的面积相等. (12 分) 5

26. (本题满分 14 分) 解: (1) A(0, ,B (4, ······························································································ (2 分) 2) 0) 设直线 AB 的解析式 y = kx + b ,则有

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?b = 2 ? ?4k + b = 0
1 ? ?k = ? 解得 ? 2 ?b = 2 ?

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1 ∴ 直线 AB 的解析式为 y = ? x + 2 ············································································ (3 分) 2 (2)i)①点 E 在原点和 x 轴正半轴上时,重叠部分是 △CDE .
则 S△CDE =

1 1 1 ? 1 ? CE CD = BC CD = (4 ? x) ? ? x + 2 ? · · 2 2 2 ? 2 ?

=

1 2 x ? 2x + 4 4

1 BO = 2 ∴ 2 ≤ x < 4 ······················································· (4 分) 2 ②当 E 在 x 轴的负半轴上时,设 DE 与 y 轴交于点 F ,则重叠部分为梯形 CDFO . Q△OFE ∽△OAB OF OA 1 1 ∴ = = , ∴ OF = OE OE OB 2 2 又Q OE = 4 ? 2 x 1 ∴ OF = (4 ? 2 x) = 2 ? x 2
当 E 与 O 重合时, CE =

x ? 3 ? 1 ?? ∴ S四边形CDFO = · 2 ? x + ? ? x + 2 ? ? = ? x 2 + 2 x ················································ (5 分) ? 2 ? 4 ? 2 ??
当点 C 与点 O 重合时,点 C 的坐标为 (0, 0)

∴ 0 < x < 2 ····················································································································· (6 分)

?1 2 ? 4 x ? 2x + 4 ? 综合 ①② 得 S = ? ?? 3 x 2 + 2 x ? 4 ?
ii) ① 当 2 ≤ x < 4 时, S =

(2 ≤ x < 4)
··························································· (7 分)

(0 < x < 2)

1 2 1 x ? 2 x + 4 = ( x ? 2) 2 ∴ 对称轴是 x = 4 4 4 Q 抛物线开口向上,∴ 在 2 ≤ x < 4 中, S 随 x 的增大而减小 1 ∴ 当 x = 2 时, S 的最大值= × (2 ? 4)2 = 1 ······························································ (8 分) 4

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3 3? 4? 4 ② 当 0 < x < 2 时, S = ? x 2 + 2 x = ? ? x ? ? + 4 4? 3? 3
∴ 对称轴是 x =
2

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4 3

Q 抛物线开口向下 4 4 ∴ 当 x = 时, S 有最大值为 ··················································································· (9 分) 3 3 4 4 综合 ①② 当 x = 时, S 有最大值为 ···································································· (10 分) 3 3
iii)存在,点 C 的坐标为 ? ,? 和 ? ,? ·································································· (14 分) 0 0 附:详解: ① 当 △ ADE 以点 A 为直角顶点时,作 AE ⊥ AB 交 x 轴负半轴于点 E ,

?3 ?2

? ?

?5 ?2

? ?

Q△ AOE ∽△BOA EO AO 1 ∴ = = AO BO 2 Q AO = 2 ∴ EO = 1 ∴ 点 E 坐标为( ?1 ,0)

?3 ? ∴ 点 C 的坐标为 ? ,? 0 ?2 ?
② 当 △ ADE 以点 E 为直角顶点时 同样有 △ AOE ∽△BOA OE OA 1 = = AO BO 2 ∴ EO = 1 ∴ E (1, 0)

?5 ? ∴ 点 C 的坐标 ? ,? 0 ?2 ?
综合①②知满足条件的坐标有 ? ,? 和 ? ,? . 0 0 以上仅提供本试题的一种解法或解题思路,若有不同解法请参照评分标准予以评分.

?3 ?2

? ?

?5 ?2

? ?

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