北京市海淀区2025-2026学年七年级上学期期末数学仿真模拟试卷

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2025-2026学年北京市海淀区七年级(上)期末数学仿真模拟试卷

一、单选题(本题共30分,每小题3分)

1.下列图形经过折叠不能围成棱柱的是(   )

A.@@@59524e0a-0325-4f46-9798-b3d91393af3e B.@@@9db66fe2-0a07-4252-b2e9-4d614ddad93e C.@@@1e102c88-fa80-48cb-8a30-0048753ed5c5 D.@@@4fd5e685-ba60-4d67-bd59-f5a598262119

2.2 026的相反数是(  )

A.eqId9123f654a5019e8ccc92f853a0faa9ca B.eqIdbfbacb056c2f94379c5f552638d9058d C.eqId93e5ce4310f1189926291f106c971217 D.eqIdb8c0b89accf1f297e678a16915c15f93

3.eqIded0d9046c1ef6a3f23483acd6ea03133年eqIdc79c562343bd2362a979662d3865020c月天问一号成功观测到星际天体阿特拉斯,观测时距离该天体约eqId8a19abe6dbc2f65ac6e3eb2a2eb394ea万千米,用科学记数法表示约eqId8a19abe6dbc2f65ac6e3eb2a2eb394ea万为(    )

A.eqId0131f0f99507930c24d4621d9ddcf1e9 B.eqIdcd575a2229688170e7af6d551eb326af C.eqId805f1572ef236a01de00982c69130bcf D.eqIdd1c4393714cefab00f8e53994746f524

4.下列结论中,正确的是(    )

A.eqId487eeb126d1e0773c5ea6406f3ffc248的次数为5B.eqId7f2b17c1060e677a710aedf72d9927ee是三次二项式C.eqId5625877f46cc3574197cca644266a55c是整式 D.eqIdf8bf6e068ee02e797414e958172f8534的系数是3,次数是2

5.已知等式eqIddeac0b916c44e6e042bda03e08181bf1,则下列等式中不一定成立的是(    )

A.eqId9faeca1ca288556ce44ea8600a04af11   B.eqIdc1cd63acd2d2ba7dca5351dcd3e50209   C.eqId081e953eac05588a62f7e2230db428e9   D.eqId735d4ab7f32c997527e5a7422d976a2b

@@@c29d53cd-f205-4354-af1b-576a9a75876a6.已知线段eqIde1cd4a1c8087ef697d82b321a3331211,点C是直线eqIdf52a58fbaf4fea03567e88a9f0f6e37e上的一点,eqId9547f6b5c7c54d216ecea72becb7d373.若M是eqId60ef95894ceebaf236170e8832dcf7e3的中点,N是eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的中点,则线段eqId411461db15ee8086332c531e086c40c7的长度为(    )

A.eqId9d1b7aa6e3693af226ef7324c37534c1 B.eqIdf3a9eed64d225267a58cd001db67e2a3 C.eqId976260cbf5e30856d4fd37a4b0a671a7  D.eqIdf3a9eed64d225267a58cd001db67e2a3或eqId976260cbf5e30856d4fd37a4b0a671a7

7.如图,点B在点O的北偏东eqId137fcc582043c37e277fb685f0a7bb2a方向上,eqId70cb1ff41b1197e15f55f2e1e7f29f6c,则点C在点O的(  ).

A.西偏北eqIdc3f2a3e1152939543973365fce025699方向上B.北偏西eqIdc3f2a3e1152939543973365fce025699方向上C.西偏北eqIdff8fb92a169c3f2ef9b3d614f8f301a8方向上 D.北偏西eqIdff8fb92a169c3f2ef9b3d614f8f301a8方向上

8.一列匀速前进的火车从进入eqIde3fcfe66444a7a05e8976fa37647b0fd长的隧道到完全通过隧道经历eqId2137b2fc62921acdadda4c580c72f6e8,隧道顶部安装了一台固定的激光发射器,它会持续发出一道垂直向下的极细激光(激光线不移动),激光照射在车身上的时间为eqIde2ca155b0b6022716c49570e7ecb4dd0,则这列火车的长为(   )

A.eqId615d4e3c6b6893ac19a0ea80c815811f B.eqIde3fcfe66444a7a05e8976fa37647b0fd C.eqId696471b98c4fa2944ecb2ccaccc05982 D.eqId8970d66a28d224a904c21d6b71a75f78

9.下列计算正确的是(   )

A.eqId91ebcfc9ca65bb98964beae7e72e4208 B.eqIdecfbff16eb99739045d6b36708d863f1 C.eqId24b0daf4943dcc6956dbbe3369083c40 D.eqId7f49db7f93bbb46c493c6695095748d7

10.小西用大小相同的正方形按如图所示的规律拼图案,其中第①个图案中有1个正方形,第②个图案中有3个正方形,第③个图案中有7个正方形,第④个图案中有13个正方形…… 按此规律,则第10个图案中正方形的个数是(    )

@@@c33641f7-069a-4bd4-9e4b-33b5ea956df7  

A.88 B.89 C.90 D.91

@@@5a664314-aa81-4675-8ed2-84dd80c641d2二、填空题(本题共18分,每小题3分)

11.比较大小:eqId602baac86c2b1668ecdfadc8a5948885     eqId5795a2f014de7b77d514d110744c1984(填“eqId392cdb9d30684cce244bef94b8d861b9”,“eqId4ff7942da6c3fc4005256fb1458557c0”或“eqId6706fe00b4e231e62d9ecbec567d526b”).

12.已知关于x的方程eqId41155b4e2684075453c6d35a585b66ab是一元一次方程,则k的值为      .

13.有两块直角三角板按如图所示放置.已知:eqId3286a091021daf0d663b4773d6237a5d,eqIdd78a2303e97a9e2f14db77c363ea3846,则eqId3cc310194575262fb45352c2fbc19868            °.

14.若一个角的补角是它的余角的eqId5ca7d1107389675d32b56ec097464c14倍,则这个角的度数为      .

15.有理数eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1,eqId071a7e733d466949ac935b4b8ee8d183在数轴上的位置如图所示,则代数式eqIdee7eff5ae4d4513f19c45fca95f738fe化简后的结果为        .

@@@33dc276b-dcc5-47d1-8e5a-bcea66cc7820

16.某商店将某种电视按进价提高eqIdf5c546b5b240133085bce442db1abdbb,然后打出“九折酬宾,外送50元出租车费”的广告,结果每台电视仍获利208元,那么每台电视的进价是      元.

三、解答题(本题共.52分,第17题6分,第18题6分,第19-22题,每小题6分,第23题5分,第24-25题,每小题6分,第26题7分)解答应写出文字说明、演算步骤或证明过程。

17(6分).计算:(1)eqIdd3fecab2a518dac2be46dd508e58fe17;(2)eqId932f5a7b0732637b26b0399167e3e5dc.

 

 

18(6分).解方程:(1)eqIdf6ad40b9a88d8b7458c44c073d6d0159;(2)eqId81b3d580c8f9c3266cbf0021e2f20cd8.

 

 

19(6分).已知eqId83347b9cfada96776f7f8ae969984f98.(1)求eqIde13c8573cf2fff14ab3c22a56536d8e1;page number 0

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(2)若x,y满足eqId7b148665bde3e7bf39d24f49d074a23e,求eqIde13c8573cf2fff14ab3c22a56536d8e1的值.

@@@d9975cb3-0489-4812-aeb6-53651daa016020(6分).根据题意画图:

(1)连接eqId73b3c032441543354c154ee67d744abb,相交于点O;

(2)延长线段eqId03902478df1a55bc99703210bccab910,与线段eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的延长线交于点E;

(3)反向延长线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e,与线段eqId9d78abbad68bbbf12af10cd40ef4c353的延长线交于点F.

21(6分).如图,eqIde52586ca2a3b783bc8092415e2d4bf6d两点在数轴上分别表示有理数eqId663a61ad241d5d874c9a9362f0ee917c,且eqIdcdd76ebfb21bb2bac714bbed2b7bb5d1,点O为原点,点C在数轴上eqIdfb8134c0bf1418693912b362cc71b00d两点之间,且eqId85bdcc2742be52cc80ba7dbe79f9a574.

@@@429ef625-3dd6-4ac3-90dc-8f37a4a3ffb5

(1)直接写出eqId380bbacf854e30e2e747fc286d2b9997__________,eqId5ccd4162c7d09f970cb77cadacdbe521__________,点C所对应的数是__________;

(2)动点P从点C出发,以每秒1个单位长度的速度向左运动,同时动点Q从点B出发,以每秒2个单位长度的速度向左运动,运动时间为t秒,当点P与点Q的距离是5个单位长度时,求t的值;

 

 

22(6分).如图,点C为线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e上一点,eqId60ef95894ceebaf236170e8832dcf7e3与eqId6b1bd1adfe4cc6566218f19970c2fd3b长度之比为eqIdcf1484ddc948caf40e4c3243bc959584,D为线段eqId60ef95894ceebaf236170e8832dcf7e3中点.

@@@b94da17d-8c0b-4f1d-a01b-07c813350551

(1)若eqId01b1927db96d84f27197b5a654b622f6,求eqIdd40b319212a7e7528b053e1c7097e966的长.

(2)点E为线段eqIdd40b319212a7e7528b053e1c7097e966的中点,若eqId8bb6f7f8126875d984f1253c7c9391b1,求eqIdf52a58fbaf4fea03567e88a9f0f6e37e的长(用含m的代数式表示).

 

 

23(5分).如图,O是直线eqId4eedae8d316c76e3d0b451256de03fb9上一点,以O为顶点作eqId20ccc37b189fa2cbc269ca0b233dac37,且eqIdef4113c492885ba7c47fe42ac792578f,eqIdb90e0f35eda1a729fed485f83da5ea9d位于直线eqId4eedae8d316c76e3d0b451256de03fb9两侧,eqIdb90e0f35eda1a729fed485f83da5ea9d平分eqIded66570f697459aba9daa2eaf9460bcb.

@@@80ee0720-c1c9-4d50-a9d9-9cb3d48220d4

(1)当eqId79684a6e92297749c005e2b23cac9710时,求eqId4cbee6d5731d6f4bb1e4f2ba3d89b7dc的度数;

(2)请你猜想eqId9fed1f57a835af4e9022e27603d12d31和eqId4cbee6d5731d6f4bb1e4f2ba3d89b7dc的数量关系,并说明理由.

 

24(6分).在手工制作课上,老师组织七年级eqIdbdaa19de263700a15fcf213d64a8cd57班的学生用硬纸制作圆柱形茶叶筒eqIdc90282d4a37c9a20620d4bbb0c263cae七年级eqIdbdaa19de263700a15fcf213d64a8cd57班共有学生eqId0e6f1af4b44b2e97e8f319bab4ae9010人,其中男生人数比女生人数少eqId61128ab996360a038e6e64d82fcba004人,并且每名学生每小时剪筒身eqId4b72ac611ae66b86761e080761d9aabc个或剪筒底eqIdcfd121f4aeca8a78a320fca1fd1c4d07个.

(1)七年级eqIdbdaa19de263700a15fcf213d64a8cd57班有男生、女生各多少人?

(2)要求一个筒身配两个筒底,那么如何进行人员调配,才能使每小时剪出的筒身与筒底刚好配套?

 

 

25(6分).为了迎接新学期,书店计划购进eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d两类书刊,eqId5963abe8f421bd99a2aaa94831a951e9类书刊和eqId7f9e8449aad35c5d840a3395ea86df6d类书刊的售价分别是eqId58b184c94e38f1e5dbe750b2168c2a37元eqId4333bb203d02814c146ed587b69ea69d本和eqId4b7f27ebcef70a3ebbbe8d2e53ea0896元eqId4333bb203d02814c146ed587b69ea69d本,且每本eqId7f9e8449aad35c5d840a3395ea86df6d类书刊的进价比每本eqId5963abe8f421bd99a2aaa94831a951e9类书刊贵eqId61128ab996360a038e6e64d82fcba004元eqIdc90282d4a37c9a20620d4bbb0c263cae已知购买eqId6b685c556cc423e4833c1dc671a134cc本eqId5963abe8f421bd99a2aaa94831a951e9类书刊和eqId90dd550e1ad9bbf01687ffb4aab788ec本eqId7f9e8449aad35c5d840a3395ea86df6d类书刊共需要eqIdb6b058d8135a950418f948bdf378efb3元.

(1)每本eqId5963abe8f421bd99a2aaa94831a951e9类书刊、eqId7f9e8449aad35c5d840a3395ea86df6d类书刊的进价各是多少元eqId82bbee662e242611afdbdae4b8a36a7c

(2)若该书店第一次购进eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d两类书刊共eqId98ac61206cb12cf6686bb0facf635010本,全部售完后总利润为eqId1e9100a6b883eea24fba295bb54ec3aa元,则该书店第一次分别购进eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d两类书刊各多少本eqId82bbee662e242611afdbdae4b8a36a7c

(3)若第二次购进同样数量的两类书刊,且两类书刊的进价都比上次优惠了eqIdf733b1ceeead9ff892539d46a23f3626,再次销售时eqId5963abe8f421bd99a2aaa94831a951e9类书刊售价不变,eqId7f9e8449aad35c5d840a3395ea86df6d类书刊打折出售,全部售完后总利润比上次还多eqIdb53c7539ed297ea63b9ace6f5cc58ca8元,则eqId7f9e8449aad35c5d840a3395ea86df6d类书刊打了几折eqId82bbee662e242611afdbdae4b8a36a7c

 

 

 

 

26(7分).背景知识:数轴是初中数学的一个重要工具,利用数轴可以将数与形完美地结合,这种解决问题的思想叫做数形结合思想eqIdc90282d4a37c9a20620d4bbb0c263cae已知,点eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d是数轴上不重合的两个点,且点eqId5963abe8f421bd99a2aaa94831a951e9在点eqId7f9e8449aad35c5d840a3395ea86df6d的左边,点eqIdac047e91852b91af639feec23a9598b2是线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e的中点,点eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d,eqIdac047e91852b91af639feec23a9598b2分别表示数eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1,eqId3e1a69049bf1f56e80017338cc1d4158请回答下列问题.

【特例感知】(1)若eqId8e258ab9e600435b37465092243d99f6,eqId019eb94a6a2b38308811470d860e1a20,则eqId306131877e7eec647a63edb9711f07c6______,eqId81dea63b8ce3e51adf66cf7b9982a248表示的数为______;

【规律探究】(2)如图,利用数轴思考探究,点eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d之间的距离表示为______,eqId81dea63b8ce3e51adf66cf7b9982a248表示的数为______eqIdfd995178601c2ad7b40f973d268c7bb7用含eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1的式子表示eqId04582116cd765fcc5a52f44279ad6c94;

@@@17b6f015-b97f-4726-9238-187b42fddfcb

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参考答案

题号

1

2

3

4

5

6

7

8

9

10

答案

D

A

B

C

C

B

B

C

C

D

 

11.eqId392cdb9d30684cce244bef94b8d861b9

12.1

13.54

14.eqId79a97bb4dcfab4ec7539bc783d563c49

15.eqIdf5021f05814ea7244a487cd87648b5af

16.1200

17.(1)解:原式eqId794c77b590ef7d9755c6de245e3c3b4c;

(2)解:原式eqIdd2ea33961b16bfcc9ebd48a8ee3764bd

eqIdbd938ebfba774aeb803f2844807bab6d

eqId0188f45fa33feb2ecc908619d50989e4.

18(1)解:eqIdf6ad40b9a88d8b7458c44c073d6d0159,

去括号得:eqId298aadf740b6f495998885799e262e99,

移项合并得:eqIddcecb03dd1700a29300bd8330939fe22,

系数化为1得:eqIdb650820d7bed48ed67a2869ad8c65ff1;

(2)解:eqId81b3d580c8f9c3266cbf0021e2f20cd8,

去分母得:eqIdbfb036329b639f21227424d03795525d,

去括号得:eqIdcf8052b73d00bc4d62a7008f3aaf7da6,

移项合并得:eqId5a8068f1472e337aa6016c5d3a07ce60,

系数化为1得:eqId639c3d2ff5ee566fcc1b69c65712a661.

19.(1)解:eqId43075d25169cb4d02160e6a182da7869

eqIdbee8062fd52529d19de5c14bf54d3e10;

(2)eqId7b8c1ea6173e22f203e790a74acb73ad,

eqIdd964e6685cc1113075273e3edb5d9663,

解得eqId3c8eb30ffc364d331b48097dbfab85de,            page number 2

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