15.2.2分式的加减 课后训练 2025—2026学年华东师大版八年级数学下册

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15.2.2分式的加减课后培优提升训练华东师大版2025—2026学年八年级数学下册

一、选择题

1.计算eqId2bc61267869684fed321adb776cc0da4的结果是(    )

A.eqId9c39363303ed5a1d73ef9c1148d871b2 B.eqId1e8954d89dddb533d9e62f6836a579f7 C.eqId5592534a6654524e073c7f109aad0ba9 D.eqIdb7919046b50ec77c2a1851f099da54ed

2.代数式eqId695b94998f76717f39941c35d6f8a33f的化简结果是(  )

A.eqIda8dead0f47f35151809a3216b9d27fc3 B.eqId0fbfae03d027fed4eb488568aceee662 C.eqId56d12e608b5905a71b10b756b5f5343c D.eqId43d4ce025d83ffcbe75fd008d246f524

3.如果eqId0ae78b4e16cfb3c06c95b8779a23585f,那么代数eqId9664ce95d1eed211c20d4abe75c06fdb的值为(   )

A.eqIda7ffe8515ff6183c1c7775dc6f94bdb8 B.eqIde675cc8dca9bb36e8240d19cb4fc53ac C.eqId6b43bdce2e29da5566a06ed328f387d6 D.eqIdf38a44c240455c0366acf401e289a5f2

4.现有一列数:eqIde72adb45c60c2f63b46e65ff787302bf,eqId3e88093a749c0d46e0ee931ecfaff925,eqId6c1ccc6c74b8754e9bcbb3f39a11b6f1,eqIddaf464629fa321a6ff7401ab79f07083,…,eqIda7230de53663c75658c58bbf206a0085,eqId96abfe2da27a63e6affb19a0c80236d9(n为正整数),规定eqId039e4fe671d61e59b96ee525c9df43e8,eqId1df7f2beffd703583d4b4db720bf778b,eqId4ea11bd18577ee314988bc70b0caf23f,…,eqId667bf06ddd83d9da7f2d0e01cc3319b1,若eqId37d6e8a35c5a4f76bb3791261681d318,则x的值为(    )

A.1 B.1.5 C.2 D.2.5

5.若eqId289295fad0d1c11fc74470e8bb0c6be0,eqId87a655d6935ae3f646e17ff72bc213e2,则eqId86c5b46e81044efa7090be7ae678ff72的值是(    )

A.6 B.7 C.4 D.eqIdec85f29c0860b57a8f0cf8098c13a97e

6.已知eqId4a7dbc702617c765a573961953cc0901,则eqId11975d3bcd4cfd97bddfe86f5b213798的值为(    )

A.eqIdf89eef3148f2d4d09379767b4af69132 B.2 C.4 D.eqId56d266a04f3dc7483eddbc26c5e487db

7.已知eqId3db3fd4869bced363ddf4b50db7689e1,则eqId35dca2002ff7fcf6b6786f78a1a8c187的值为(    )

A.eqIdd297eab7380f6a28ec010218d9ab4ba1 B.eqIdec85f29c0860b57a8f0cf8098c13a97e C.5 D.eqIdf89eef3148f2d4d09379767b4af69132

8.已知eqId11a4ad5c964b93af75de44f9eea03a4f,eqId916dd24222dc11b50f29ab787b5d3c38,则(    )

A.eqId00856b4c9bb6c037b5ac78cc408ab4b7 B.eqId927ebb016c082fc9a95bd21018d31802 C.eqId2ab662b3ce6fca25065bb811d8861729 D.eqIdb3c4f8faea2797ee88cbe6eb57be68b4

二、填空题

9.已知eqId61f97784c9b4b91bd008cb52009a4896,则eqId3949835617c8e239c7955da02aad7ae3的值为_______ .

10.若eqId3e4a2c03eee3e199949bb9426365caf9,则eqIdb1f9fabbbe61a759e52ec975215e2e7c________.

11.如果eqId64be2ec55730afd47db77fdd156788e1,那么代数式eqId8d73156762ac3dab65c8c8f597afe6ff的值是____________.page number 0

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12.利用倒数性质可以解决一些问题,已知:eqId9a742891dc364ebe3fcff9e138a52e20,易知eqId38f0e9c04402a0ffdaa25c3e3c82c7dd,取倒数得eqId7a0feb33f37204a2069eb36e776eed2d,化简为eqId5676f7b5b8e00dc101dd5850806e0fb6,则eqIdb635e0a4190ca67570898bed5da2dad2的值为______.

三、解答题

13.先化简,再求值:eqId1b29c646881e9c3935fe29d5d4aeeb7b,然后从0,1,2中选取一个合适的数,求式子的值.

 

 

 

 

 

 

 

14.化简:

(1)eqId11543069ee2e1d9a38c43b9ccbf47271

(2)eqIdce650ffc32c56c6ef03af1cc4d97205c

 

 

 

 

 

 

 

15.如果两个分式eqIdac047e91852b91af639feec23a9598b2与eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3的差为整数eqId0a6936d370d6a238a608ca56f87198de,那么称eqIdac047e91852b91af639feec23a9598b2为eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3的“模范整分式”,整数eqId0a6936d370d6a238a608ca56f87198de称为“模范整值”

如: eqId038fef4bbb3bc93ca15e99c11b85a9e9 则eqIdac047e91852b91af639feec23a9598b2为eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3的“模范整分式”,“模范整值”eqId89d500af0ca164f4f04b67a080ba6189.

(1)已知分式 eqId3c57b7b67a376b8bcbeb9487a2f64276.判断eqId5963abe8f421bd99a2aaa94831a951e9是否为eqId7f9e8449aad35c5d840a3395ea86df6d的“模范整分式”,若不是,说明理由;若是,请求出“模范整值”eqId0a6936d370d6a238a608ca56f87198de;

(2)已知分式 eqId0c5ad0117f06aa6f9e0f2df08694ed87其中eqId2a30f3a8b673cc28bd90c50cf1a35281为多项式,且eqIdc5db41a1f31d6baee7c69990811edb9f为eqId8455657dde27aabe6adb7b188e031c11的“模范整分式”且“模范整值”eqIdb9af6b8933cb0bcc983a15c903b5892e,求eqId2a30f3a8b673cc28bd90c50cf1a35281所表示的多项式.

 

 

 

 

 

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16.在初中数学学习阶段,我们常常会利用一些变形技巧来简化式子,解答问题.

材料一:在解决某些分式问题时,倒数法是常用的变形技巧之一,所谓倒数法,即把式子变成其倒数形式,从而运用约分化简,以达到计算目的.

例:已知:eqIdb08635365ea501bff737d957c24d9bb5,求代数式eqId13f40c24c64bbb0645fcf585f4e66872的值.

解:∵eqIdb08635365ea501bff737d957c24d9bb5,∴eqIdf3a035b607d843cdfae8a371dc1569ce即eqIdeb81872d440b75836121a1ffff39c2b2,∴eqId00b5da5a8e02d9b8c2e5396e4aacdda8

材料二:在解决某些连等式问题时,通常可以引入参数“eqIdf0a532e15e232cb4b99a8d4d07c89575”,将连等式变成几个值为eqIdf0a532e15e232cb4b99a8d4d07c89575的等式,这样就可以通过适当变形解决问题.

例:若eqId184e508a3f4f490ab4a9084ed9356868,且eqId87ad1416311af625004c330c5ec3b1f1,求eqId4b285de18365ca8b0cf6b0cfe564de10的值.

解:令eqId828217ff3282774e5cc42c759b2b300d则eqId540ed28e991b7bc2f30f68ce9811f5fb,eqId611a767c9ec390a4c5ea8b1bc62dca7d,eqIdf3597cee90e17c9d975ef49a7f6e10f9,∴eqId4ff3703d1e4f4e7140a6070e4d629686,

根据材料回答问题:

(1)已知eqIdc650cb8cc960d4009033a461841f5e77,求eqIddd43ab16b2e3c663eed3f38e1cc8b26e的值;

(2)已知eqIdaa71db07ef6b2972478f61d843e1eb89,求eqIdccc32ce5a2b12e48f8a4707ecb3b057b的值.

(3)若eqIdaec737627ce8b62833376dea4160d8e9,eqId38f0e9c04402a0ffdaa25c3e3c82c7dd,eqId1c412d5329ba909164329663b7eecdfe,eqId8a23eaadf099e3ba43b0ab0e0e01d843,且eqId56667aabbe787eb1c3189d487d203e22,求eqId3e823000035b4482d4868da22c0f9fc7的值.

 

 

 

17.操作发现:阅读下列解题过程:已知eqId9a742891dc364ebe3fcff9e138a52e20,求eqIdb635e0a4190ca67570898bed5da2dad2的值.

解:由eqId9a742891dc364ebe3fcff9e138a52e20知eqId38f0e9c04402a0ffdaa25c3e3c82c7dd,所以eqId7a0feb33f37204a2069eb36e776eed2d,即eqId5676f7b5b8e00dc101dd5850806e0fb6.

eqId9dcd875f691b6fff3e16e5afa96c0069,eqId998f5dfdf4d4abca18c162f524d6f047的值为7的倒数,即eqId3f766c6de960a7c1292ce9b9604475f8.

以上解法中先将已知等式的两边“取倒数”,然后求出待求式子倒数的值,我们把此题的这种解法叫做“倒数法”,

(1)已知eqId396fcef5b1d99a74bb395aaf856573e1,求eqIdb635e0a4190ca67570898bed5da2dad2的值;

(2)实践探索:请你利用“倒数法”解决下面问题:

已知eqIdded2ca94896381c10d236b73478c5825,求eqId76480abf9baf345ac1a16fb53b88f7ce的值;

(3)问题解决:

已知:eqIdd84e85eae95d48b791fbc3004c111c29,eqId19f28628d11f332a594c69f6047e5256,eqId0dbfa094091b846e3c6c15b5608777a4,求代数式eqIde1433c8103033c67232f2f9ae189608d的值.

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