2026年初一数学期中复习(2)

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2026年初一数学期中复习(2)

一、单选题

1.下列说法正确的是(     )

A.eqId395addb4439331615a7f220289d8b62c的平方根是eqId72b8bee40319fe80e512d221cfe252a7         B.eqId8bf3d3564c61e5e9c39a9e2cf2de048b的立方根是eqId72b8bee40319fe80e512d221cfe252a7

C.eqId39f3fae1304b36c0aa24d98d4ae0508b的平方根是eqId2dcee5c0f947e3b4514661d46f77d170

D.算术平方根等于本身的数是0、1

2.数轴上表示eqId586097068c87108c7cbd17f5223769c5的点eqId5963abe8f421bd99a2aaa94831a951e9的位置应在(     )

A.1与2之间      B.2与3之间 

C.3与4之间      D.4与5之间

3.如图,将△ABC沿eqIdf52a58fbaf4fea03567e88a9f0f6e37e方向平移eqId61128ab996360a038e6e64d82fcba004eqId9efa9fbcfb9595e2f031aa691db4564b得到eqId72cb97395ebc5ee1b212afb7a97b985c,已知△ABC的周长为eqId5f573bd1591735843cc5852535af976feqId9efa9fbcfb9595e2f031aa691db4564b,则四边形eqId1507da3daed983c2f355d4caebb66d72的周长为(     )

@@@2ca5684a-6a15-4bbb-afbc-3c06c347f41a

A.eqIdadf1a2d369203cbb461d38388e44ea08eqId9efa9fbcfb9595e2f031aa691db4564b    B.eqId51ed533272ac9d01f1ce3e741b034d4feqId9efa9fbcfb9595e2f031aa691db4564b    C.eqId5f573bd1591735843cc5852535af976feqId9efa9fbcfb9595e2f031aa691db4564b  D.eqIdf179a5e46197a7f5f5646fefd47c1989eqId9efa9fbcfb9595e2f031aa691db4564b

4.如图,将线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e平移到eqId9e52a8f07834cbbbe4224962672fbbb2,已知三个端点的坐标eqId801c5b117e9efbcf853e78fc422c5979,eqId81dbacadb06d62c93b01789abc5194e9,eqId3c51f2fdbdf523df39385baa5fdc17a8,那么第四个端点eqId8455657dde27aabe6adb7b188e031c11的坐标是(     )

@@@0586448f-cf40-4f64-bf9b-d9d20f9eff83

A.eqId33d1b3d723539d387df60646d5ef9f55     B.eqId6b4bab5e9512cf8aed1308723b90a1b7      C.eqId1103238cea507c3bc3f98a96eef5433f  D.eqIda6314f43ef938889856805d83bd9b7bd

5.已知线段eqId6609cc372831939ad4beb3770b0b0ca8,eqId411461db15ee8086332c531e086c40c7eqIdd58f4fa90e6a54c669aacf8f0b43caa0轴,若点M坐标为eqId27cf818dd484cc4cebd40a5f28eb8e9f,则N点坐标为(     )

A.eqIda9a98c743bf6211bd15150093d7729e5        B.eqIda9a98c743bf6211bd15150093d7729e5或eqId081cd41dab0f2a8f84b0e9f1df4843fb 

C.eqId081cd41dab0f2a8f84b0e9f1df4843fb D.eqIda77b96fc51eb8a9d03ced254ce8b78be或eqId6d0d32ea3bbad35e0af9ecff1c62d044

二、填空题

6.如图,将直尺与eqId3c030b25575d683af91c06e6a3e4f463角的三角尺叠放在一起,若eqId90024c840a5a166edb0cad21f47d874e,则eqId84123d56994a7416790420fedd81b745__________度.

@@@d040ce5a-f81e-4de4-8ec9-f0b725135cd4

7.已知,若eqIdf52a58fbaf4fea03567e88a9f0f6e37e与eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的夹角为eqIdb975f35ecfc9641688e682bcee70e1f4,eqIdf586d178cb0f69dd900967766ba8950a,则∠2的度数为______.

@@@0f1e2fb3-690f-4d3d-a240-72260195ec8c

8.已知eqId55d2f4285604c1536f929c0162c814de,eqId47bb3f35e3db7c1f3a3dd3eb20151b5f,eqId43573249e132a8f595cf98b4e9cd5b3d,则eqId0533d03d8cd9f102563dcc5c725e4954___________.

9.已知eqIdf6d0a8865088eb8dce20cdddfc78e5e6,eqIde02890aba78b206da39b275db36d492a,eqId2499cdf1d1ae9615b6a222401c608fcc,eqId9a459b3cc80dc2fea2413680005649a6,则eqIde75f5034587de33d6cf9f5e1bf0dbd58的立方根是________.

10.比较大小:eqId3325273c2f82fa4ae56c6d85ac66b2d1______eqIdfb8f58755aee89fb2cf72ba518dcee2a.(填“eqId392cdb9d30684cce244bef94b8d861b9”“eqId4ff7942da6c3fc4005256fb1458557c0”或“eqId6706fe00b4e231e62d9ecbec567d526b”)

11.已知a是eqId35361e76a7c85d1886728c8d0200b234的整数部分,b是eqId35361e76a7c85d1886728c8d0200b234的小数部分,那么的值为   _____  .

12.如图,已知eqIddbeb52a84259fde9e9ca355fb025f43d,则点eqIdbb23814c40a4929a6483d60a413968da的坐标为    _____.

@@@74a6e814-fd01-4dd5-a2e2-a4bb50c82c56

三、解答题

13.计算:(1)eqId9a08b1b0ea1cc89d137530024c374acb;

 

 

 

 

 

(2)eqId5c30f34fb734a893cbeb549641d7b331.

 

 

 

 

 

 

14.解方程:(1)eqId4a836c7f8009e5b85389301960d96b0a;

 

 

 

 

 

 (2)eqId9972086576faee535c1314d38f559074

 

 

 

 

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15.已知一个非负数c的平方根是eqIde0c0adbbd84df799d851d0e803d09c2c与eqId60d9e2688339da9655e4e58c10863670,eqId9061bb2a84cecfde7089611c59e63be0的算术平方根是4.

(1)求a,b,c的值;(2)求eqId45cf9ac4f0c1fd80d6562dd088234693的立方根.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

16.补全解答过程.

如图,eqId1d1717b5311ed602f1730e86d9322647,eqIdd1598ad60d36f87f7fed1a58cf5c2c42.求证:eqIdc436f108fd4921dae15ecff19270237e.

@@@bc4ae110-bc40-4f28-8f0a-d29064b7efcc

证明:∵eqIdf8d11b492476926aaea82de48e28282a (邻补角的定义),eqId1d1717b5311ed602f1730e86d9322647 (已知),

∴eqIdf3e656bb343181eef14d4297b84af5dd(                ),

∴       eqId2a9bfa68259d7a331be323b2038d628a         (内错角相等,两直线平行),

∴eqIdd5324ce5fd43c5b05e8fad764880f7b3       (                            ),

∵eqIdd1598ad60d36f87f7fed1a58cf5c2c42(已知),

∴eqIdafde09d82ae96f00c135732baee64776       (           ),

∴eqIdc436f108fd4921dae15ecff19270237e(                            ).

 

17.如图,eqIdc4e6f5241f614c7d310cbf3e25a6d615,eqId1beff025b32f688316009552d2fd8e1a.

(1)试说明:eqIda710061331ee8ce2a93186f7fef49fa9;

(2)已知eqId2552be65d4bfa6011d49dae83f365a4b,求eqId4c5d3f1c7f7ffdf572978468d5a3adde的度数.

@@@808afe23-a9a8-4d3a-a3f0-88f11f16f197 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

18.△ABC在平面直角坐标系中如图所示,点A的坐标为eqId44fd3d405c93fb16ea10a879db5301bb.

(1)请求出eqIdb5cbff84327e964f912a54032e76ccc9;

(2)x轴上是否存在点P,使得eqIdca65de660e704fd59496e1a2c24fa311,若不存在,说明理由;若存在,求出P点的坐标.

@@@4e827d68-47d2-4ca3-a96b-22021aa1a1cc 

 

 

 

 

 

 

 

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《2026年初一数学期中复习(2)》参考答案

题号

1

2

3

4

5

 

 

 

 

 

答案

D

B

A

B

B

 

 

 

 

 

 

1.D

【分析】本题考查了平方根、立方根及算术平方根的概念,解题的关键是明确负数无平方根、立方根的唯一性及算术平方根的非负性.

根据平方根、立方根、算术平方根的定义,依次分析各选项的正误.

【详解】解:A、负数没有平方根,eqId3edbd40e04e2a943051fa83d6e511add不存在平方根,此选项不符合题意;

B、eqId8bf3d3564c61e5e9c39a9e2cf2de048b的立方根是eqId274a9dc37509f01c2606fb3086a46f4f(立方根唯一),不是eqId72b8bee40319fe80e512d221cfe252a7,此选项不符合题意;

C、eqId39e65eccaa7119ae2617ab8e3b4824a7,eqIdb8860d9787671b53b1ab68b3d526f5ca的平方根是eqId72b8bee40319fe80e512d221cfe252a7,不是eqId2dcee5c0f947e3b4514661d46f77d170,此选项不符合题意;

D、eqIdc95b6be4554f03bf496092f1acdfbb89的算术平方根是eqIdc95b6be4554f03bf496092f1acdfbb89,eqIdbdaa19de263700a15fcf213d64a8cd57的算术平方根是eqIdbdaa19de263700a15fcf213d64a8cd57,故算术平方根等于本身的数是eqIdc95b6be4554f03bf496092f1acdfbb89、eqIdbdaa19de263700a15fcf213d64a8cd57,此选项符合题意;

故选:D.

2.B

【分析】根据无理数的估算计算即可;

【详解】解:eqId5a5bffa3f47eeb29ff7d8bf5cd7c474e,

eqId4e7c51b7095ff0cd69000d74466d1b16,

eqId0110e792319659706c84a63404ffa08b,

eqId2de0d10ef8b748d4531250c37c5d3f9e点eqId5963abe8f421bd99a2aaa94831a951e9的位置应在eqId61128ab996360a038e6e64d82fcba004与eqId5ca7d1107389675d32b56ec097464c14之间.

3.A

【分析】此题主要考查了平移的性质,根据平移的性质得出eqId9ee2747a7ab996a97b98533c71fd14e9是解题关键.根据平移的性质得出eqIdc43f9e3d40b0dd4d42bf33dcd484f65c,eqId9ee2747a7ab996a97b98533c71fd14e9,进而得出答案.

【详解】解:eqId0a29b8e848581230a967cd4c0d091c24是由eqIde428e7a09732be85c1224e9c8f6a71c5沿eqIdb79dd200766db27fb90d6bd1992cf658方向平移eqId5992036c739557714bc05e08a93bdd3c得到的,eqIde428e7a09732be85c1224e9c8f6a71c5的周长为eqId1cfa1a62bb3b75f97a669e89569c815e,

eqId7416ecaad566268fa5ee4e0532edd1ed,eqId9ee2747a7ab996a97b98533c71fd14e9,则四边形eqIdf3632f0b1f123030f7f1bbb66dd740a6的周长为eqId40b711fb19c3ef525fbe49b18ea347d2.    

故选:A.

4.B

【分析】根据平移的规律进行求解即可.

【详解】解:由题意得,线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e平移到eqId9e52a8f07834cbbbe4224962672fbbb2的横坐标变化为:eqIdd9a895424ba3889ff048905926a42fc3(向右平移4个单位);

纵坐标变化:eqIdd4bbb5b548882397e3b4214156f9b893(向上平移2个单位),page number 2

 

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