2025年中考数学三轮复习备考 操作探究问题重点题型预测练

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操作探究问题

一、单选题

1.对任意实数eqId81dea63b8ce3e51adf66cf7b9982a248,可用eqId2c4f5908d6a1217e493ed7586b6964dd表示不超过eqId81dea63b8ce3e51adf66cf7b9982a248的最大整数,例如eqIded90daa08b15619306e7b8341c115f88,eqId6ec30acc4f5d73891eb485a64e9d5d44,若将eqId81dea63b8ce3e51adf66cf7b9982a248变换成eqId5e71038bfb2d4afe1e4e035f7053c84f称为对eqId81dea63b8ce3e51adf66cf7b9982a248进行一次操作,例如:现对54进行如下操作eqId35c14c1bb7bd8f25cd0d49b96a0265a9,这样对54进行3次操作后变为1,对一个正整数进行类似操作,下列说法正确的个数是(   )

①对37进行一次操作后的结果是6;

②对138进行两次操作后的结果是3;

③对一个正整数一直进行操作,最终得到的结果是0;

④若正整数eqId81dea63b8ce3e51adf66cf7b9982a248进行3次操作后变为1,则eqId81dea63b8ce3e51adf66cf7b9982a248的最大值是225.

A.4个 B.3个 C.2个 D.1个

2.eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqId4f0b0c632e5cbe9446ddae24d3b206fe,按照以下操作:

@@@2bb729c3-a56a-4d3c-825d-ba289540d303

第一步,分别以点eqId5963abe8f421bd99a2aaa94831a951e9和点eqId7f9e8449aad35c5d840a3395ea86df6d为圆心,大于eqId20d49404351575703cfe8325d1352ec9的长为半径作弧,两弧相交于eqIdad056c25c0fdcbcc765eb5cbc6093f2b两点,作直线eqId49b50357a6545cae8348e3059312f520分别交eqId374fe9986ebbc986fc422e514ab93a51于点eqId9c0f067a2a348ceb24a408f82992eab8;

第二步,分别以点eqId5963abe8f421bd99a2aaa94831a951e9和点eqIdc5db41a1f31d6baee7c69990811edb9f为圆心,大于eqIdc93148adbc6e856da9a9d263f485d003的长为半径作弧,两弧相交于eqId7789a500686c7a73770404ead6af0590两点,作直线eqId411461db15ee8086332c531e086c40c7分别交eqId8cca04b2a2b61d62a809776670a60c09于点eqIdf6bce3d91ca23b86d8c6625f2632e437;

第三步,连接eqId5d2692bcedef0ea12afc5ceb5e6fc957.

根据上面操作,甲、乙两位同学分别写出了一个结论.甲同学结论:eqId34a53752ef5ba71dcf348aed225ec38e;乙同学结论:eqIdd3fcbf4015e1e2a33aa55cfb4ea82dca的周长为1.8.下面说法正确的是(    )

A.甲同学结论正确,乙同学结论错误 B.甲同学结论错误,乙同学结论正确

C.甲、乙同学结论都正确 D.甲、乙同学结论都错误

3.如图,将半径为单位1的圆放在数轴上,点A与原点O重合,按下列步骤操作:①将圆沿着数轴向右滚动一周,点A到达点eqIda18722354086c42e62334983fc50eb6a处(如图①);②将圆沿着数轴向左滚动两周,点A1到达点eqIdcd3b9e816b14051f785aa5aae72b8eed处(如图②);③将圆沿着数轴向右滚动三周,点eqIdcd3b9e816b14051f785aa5aae72b8eed到达点eqId04b56e44e4f0424a2b7a45567120a2e4处;…则点eqIde645ae0b78ad4ca300e3889ca3f9bcce表page number 0


示的数是(    )

@@@e1ea5408-0bb3-4ebf-9582-413d0419f791

A.eqId6399bbcacab58d6b6a1dfaea0fd0e696  B.eqId3b149fb7b03bb2991173845c42593025  C.eqIdcec8e06747aa5d790c320c60cc142974  D.eqId3a93a37165a103a9a53cf287d689e9dc

4.题目:“如图,在正方形eqId5cb3f9a5da641be35117fd35ba07a6aa中,点eqId2a30f3a8b673cc28bd90c50cf1a35281在eqIdb79dd200766db27fb90d6bd1992cf658边上(不与点eqId5963abe8f421bd99a2aaa94831a951e9,eqId7f9e8449aad35c5d840a3395ea86df6d重合),点eqIda0ed1ec316bc54c37c4286c208f55667在eqId764509115979e9958101808383672ec0边上(不与点eqId7f9e8449aad35c5d840a3395ea86df6d,eqIdc5db41a1f31d6baee7c69990811edb9f重合),将线段eqId0215e13a9fb5574d5194aeb9507a98aa绕点eqIda0ed1ec316bc54c37c4286c208f55667顺时针旋转,使点eqId2a30f3a8b673cc28bd90c50cf1a35281的对应点eqId895dc3dc3a6606ff487a4c4863e18509落在正方形eqId5cb3f9a5da641be35117fd35ba07a6aa的边上,将线段eqIdd87448e09eaa816e50ae92d111d5ded6绕点eqId895dc3dc3a6606ff487a4c4863e18509顺时针旋转,使点eqIda0ed1ec316bc54c37c4286c208f55667的对应点eqId73465a1f9aa03481295bf6bd3c6903ac落在正方形eqId5cb3f9a5da641be35117fd35ba07a6aa的边上,eqId72cdcc58d8679574ac955c1ae5b01e57依次操作下去.若经过多次操作可得到首尾顺次相接的正eqIdb6a24198bd04c29321ae5dc5a28fe421边形,求eqIdb6a24198bd04c29321ae5dc5a28fe421的值.”对于其答案,甲答:eqId6862dd3c43647ce7d7935a7d205b54d2,乙答:eqIda50b36a56f772e8d14c8fbc32f098300或eqIdcd3304e23f3b0f9569c4140ca89b6498,则正确的是 (       )

@@@5651a3df-d5a8-4862-844d-cd4ee94c2cad

A.只有甲答的对 B.只有乙答的对

C.甲、乙答案合在一起才完整 D.甲、乙答案合在一起也不完整

5.有一列数eqId1870afb17e3f1776a2ca564035a71e77,将这列数的每个数求其相反数得到eqId36cd2052417ccb1650cc533f62273aab,再分别求与1的和的倒数,得到eqIdcdcec5307afb7fed8a2ccf0057c161b3,称eqIdcdcec5307afb7fed8a2ccf0057c161b3为一次操作,记为eqId99a58d2f2ef54bddedcb3ca40b1b43bb,第二次操作是将eqId99a58d2f2ef54bddedcb3ca40b1b43bb再进行上述操作,得到eqIdb748925e3d725b96ec083d47cf9936bd;第三次将eqIdb748925e3d725b96ec083d47cf9936bd重复上述操作,得到eqIda8d07b0f8eb07d8d2407c84e6e604615以此类推,得出下列说法中:①eqIdea8e9f20049028ab7cd06d4d4dfe8d26;②eqIda1c1d4382ad291924d8545f6680e8e0a,③eqId8e780260e07811769820a4928aef85b3,正确的有(  )个.

A.0 B.1 C.2 D.3

6.小明同学手中有一张矩形纸片eqId411b38a18046fea8e9fab1f9f9b80a5f,eqIdc9d291cfc6c4677d20f8674c57322b46,eqId5c52b5f8acda412ddceb94c4e4e6b7f0,他进行了如下操作:第一步,如图①,将矩形纸片对折,使eqId03902478df1a55bc99703210bccab910与eqId0dc5c9827dfd0be5a9c85962d6ccbfb1重合,得到折痕eqId411461db15ee8086332c531e086c40c7,将纸片展平.第二步,如图②,再一次折叠纸片,把eqId9eda61683bb1d4d62441f0625097b477沿eqIdf50b3ae183997b707d16eb4e7f6712fa折叠得到eqId47a5488833625dd1e76e2ac5f9cf855c,eqIdc9b123ae31090740589ba27a846620b5交折痕eqId411461db15ee8086332c531e086c40c7于点E,则eqIde5b3bd5e6bc2a0a277d279bb01af9584到eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的距离为(    )page number 1


@@@ad71e7f5-eb4a-4aff-8166-1220f1e5657a

A.eqId80b1fe1b971b780e443a9b13621611c5 B.eqId2e7ace5e86a041e4f2c2a75eb0d247b5 C.eqId876115a273109c1e2fb0c8a7ceadcda8 D.eqIdd2504b745775c8f6239570e2e27e3ebe

7.将有序实数对eqId4562f3225c98cf5cb11b47d98c9cc9c3进行操作后可得到一个新的有序实数对eqId188f296cb76719fde47ad086573bb8f4,将得到的新的有序实数对按上述规则继续操作下去,每得到一个新的有序实数对称为一次操作.例如,eqId5265d99095b635f62c7915298ec0e963经过一次操作后得到eqId66f96d54ce364572ca3633def94e7060经过二次操作后得到eqIdc6f57b53b4ed8a78ee9fa4833a134009.下列说法:

①若eqId4ebf9f9e6000eb9d82df1cb0d9da59e1经过三次操作后得到eqIdd38b1e00db6909315f0cfc82a1ce85a8,则eqId4561e662151983a27b3e8c8fcc857f06;

②在平面直角坐标系中将eqId4ebf9f9e6000eb9d82df1cb0d9da59e1所对应的点标记为点eqId5963abe8f421bd99a2aaa94831a951e9,将eqId4ebf9f9e6000eb9d82df1cb0d9da59e1经过二次操作、三次操作所得的有序实数对分别标记为点eqIda18722354086c42e62334983fc50eb6a,点eqIdcd3b9e816b14051f785aa5aae72b8eed,若线段eqId473913c0887bb64d386f4c02f1853452平行于eqId81dea63b8ce3e51adf66cf7b9982a248轴,则eqId0df7494ac35a541f4c009a750794f9d5的面积为1;

③若eqIdb2ddc5ebafaf72a43ce10af47add4976,则eqIdc764f6b2d1788ff898c557cad844db25经过一次操作后的结果为eqId96d60e665a951ccd82b3cc9ae3d8f3ce.其中正确的个数是(    )

A.0 B.1 C.2 D.3

8.如图,已知平行四边形纸片eqId411b38a18046fea8e9fab1f9f9b80a5f,eqId3d2c15801fee2405573677484f5dcfa4,eqId07140f277a35733d8c97577ccdd4e3ab,eqId96251f9c30d745c1e6e3b60977722069.现将纸片作如下操作:第1步,沿折痕eqId85c4bdfb0db1e31e8459df1d15f9ab55折叠纸片,使点eqId5963abe8f421bd99a2aaa94831a951e9落在eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上;第2步,再沿折痕eqId5007d4aeb035e3081b8be1dd628e7565折叠纸片,使点eqId7f9e8449aad35c5d840a3395ea86df6d与点eqIdc5db41a1f31d6baee7c69990811edb9f重合.若eqIdc697c667e951df3f11503e1f4689c008,则eqId49b50357a6545cae8348e3059312f520的长为(     )

@@@a0d6060b-d705-4530-b58f-d4d34a644520

A.1 B.eqIdcaa585b9257ed0798213a9ae9b87d291 C.eqIda15aed058396d4cddf94125073da5f4a D.eqId9e2e332957bbfeb2e0643056b9aa1920

9.已知一个分式eqId1cb795d857b16feaece0cb21ffb7123f(m为整数),对该分式的分母与分子分别加1,称为第一次操作,记为eqIdc302a58c31af97d1c62f0db31daaf22d,对eqIde72adb45c60c2f63b46e65ff787302bf的分母与分子分别加1,称为第二次操作,记为eqId002b04f7c303986a282186418979c567,……通过实际操作,下列说法正确的有(    )

①eqId16d637928650350a94905291313aa277;page number 2


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