第五章 分式与分式方程 单元练习题 2025- 2026学年北师大版八年级数学下册

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2025-2026学年山东省滕州市张汪中学第二学期单元练习卷

八年级数学第五章:分式与分式方程

一、单选题

1.把分式eqId8e94e3f1016bb2a52687e5194123cf42中的x,y都扩大3倍,则分式的值(    )

A.扩大3倍 B.缩小3倍 C.不变 D.无法确定

2.下列分式从左到右的变形正确的是(    )

A.eqIdd56c3accd6a8b404af2f6b44866977a8 B.eqIda45e5575b15bdab3948634e7b351133b C.eqId3b757eebf2a74648eac0a95ec35be7d9 D.eqId47389c68e74ea5bbecf45667956a1c5b

3.使得式子eqIda4ca420837c430fb651a78c8b203df43在实数范围内有意义的eqId81dea63b8ce3e51adf66cf7b9982a248的取值范围是(    )

A.eqId24c7145166fd79adcd3e8e22ef8882bc B.eqId24c7145166fd79adcd3e8e22ef8882bc且eqIdaf1936f3caed20fd86532aa602f808d4

C.eqId124f08242308abb1041a73231d9ad1dc且eqIdaf1936f3caed20fd86532aa602f808d4 D.eqId5fd029d239948b21a074ab5e9bab8478

4.eqIdbfbacb056c2f94379c5f552638d9058d年郑州跨年音乐会于eqIdedd466424f9fc42ff269510f1be28528年eqId8da45c443af7994a26ffa9d8894e7262月eqId34584b79ec2246f47aeed8855d2762c0日在郑州大剧院开演,音乐会票价原价有四档,其中某一档门票的价格是原价的六折,用eqIde74c6091504af98fdf10ef560ecef1fe元购买打折后该档门票的数量比用eqIdacb235e39c5dde86a0c31555d9de5ab3元购买原价票的数量多eqIdb8860d9787671b53b1ab68b3d526f5ca张,求该档门票原价为多少元.设该档门票原价为eqId81dea63b8ce3e51adf66cf7b9982a248元,则可列方程(   )

A.eqId27ccc179fca7f1ef615df1cdd3283776 B.eqId0e99528d64dbe4c73be0b5702feb8187

C.eqIdffa45517b877d11b1f0f1ae922754dbb D.eqIddd2a53bbd45b5a3ba3ea4e753c5fdce8

5.已知eqIdcb1fee4981cb35db72ae4bf3ffb67c48其中A,B为常数,则eqIdf94415cec2bc8b6779491a889de09f31的值为(   )

A.7 B.9 C.13 D.5

6.下列说法正确的是(   )

A.当eqIdabc335ee14fc0b1130900cb82bcb3061时,分式eqId4ccfc5ac2d10da880c1d5783c2f38969有意义 B.分式eqIdce122e06248519df8c48cb52bad0ad88与eqId84fbee992f7edfb5ca00e9f9db9d4717的最简公分母是eqIdb5742ff250bf8a816f1e3e5760a70702

C.无论x为何值,eqIdbe9116b2015677ff5aea72b94c221ddf的值总为正数 D.当分式eqId17be0722ca4c0984ce223ebacf9c4495值为0时,eqId0b2e922ad093b97d7479cf2ad3420180

7.已知eqId5a3cd596876cd0801a11c2ab55110e84,则eqId34ba365ef11b9a4b44cfc1e6cdda2ade的值为(   )

A.eqIdd33adb74906403b0b00fcbd9fa691d8b B.eqId7294f5ae2a24ff42e84cd9773b2a7287 C.eqId0fd7b7834f33ed54661f2ce4328f661a D.eqId1d86ab7c97cd8a0b15ba5efc1be94230

8.关于eqId81dea63b8ce3e51adf66cf7b9982a248的分式方程eqIdf34e79a53934f52ee0745baa42565d72无解,则eqId294f5ba74cdf695fc9a8a8e52f421328的值为()

A.eqIdacbc6a613224461ade69362d46550474或eqId61128ab996360a038e6e64d82fcba004 B.eqIdbdaa19de263700a15fcf213d64a8cd57或eqId61128ab996360a038e6e64d82fcba004 C.eqId61128ab996360a038e6e64d82fcba004或eqId5ca7d1107389675d32b56ec097464c14 D.eqId61128ab996360a038e6e64d82fcba004page number 0

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9.李老师在多媒体上展示了一个关于eqId81dea63b8ce3e51adf66cf7b9982a248的方程eqId1bfe3bfb3221834d5205edc0f20fdf8f,甲、乙、丙同学分别提出了自己的结论:甲:当eqIda882037b9ce104ecc496e0f31a139361时,此方程的解为eqId9b384412acba251d87902ab928902f16;乙:若此方程有增根,则eqIdfcbd5bb726a08c308b48373afebbb768;丙:当此方程的解是非负数时,eqIdf0a532e15e232cb4b99a8d4d07c89575的取值范围是eqIdfbeaed9ec21e090defafcfeefe0059c5.下列判断正确的是(   ).

A.甲、乙对,丙错 B.甲、丙对,乙错

C.乙、丙对,甲错 D.甲、乙、丙都对

10.在计算分式eqId3a330c8d918f22c011c629419eb6fa88的值时,若x分别取2026,2025,2024,…,2,1,0,1,eqIdf89eef3148f2d4d09379767b4af69132,eqId4dac452fbb5ef6dd653e7fbbef639484,…,eqId2768c27ea3e501dae2b0f636cdde5cec,eqId1198d4ef1c56336c05b97a1e27d3dcc2,eqId93e5ce4310f1189926291f106c971217,再将所得结果相加之和等于(    )

A.eqIdacbc6a613224461ade69362d46550474 B.2026 C.2027 D.eqIdd8d6790ab548bc2f146c7554586e6865

二、填空题

11.计算:eqId9cb95d2126a206e729a8121268647110______.

12.已知eqId69fa88c6067600f8f279243972cd791b,eqId58b0ba5c937b27a27b5ef5ab9e51a935,则eqId9e0541697f9fffb316df0f64bf409847的值为________.

13.已知分式eqId827f4299daef7b170e0de288c0022974的值为整数,若eqId0a6936d370d6a238a608ca56f87198de是非负整数,则eqId0a6936d370d6a238a608ca56f87198de的值是_____.

14.观察下列各式:eqIde651d23dd38f769a9bffd3d6e4bf5dad,eqIda0ad1ab457736bf163b7a9a4720e4b92,eqIdbd4fa4403202b6db858f97c616b0861a请你找出其中规律,并将第eqIdd2bfe05ef6780e5f032ca3a7c96811a6个等式写出来______.

15.对于实数m,n,定义两种新运算:eqIde47e041f8d508613fa9d99ac1cd80240,eqIdafd4e01fa5f05a5c2c786fe6f57fbdf4,则eqIdec3f9997a976855621217cfa2bd15d3f的值为______.

16.对于正数eqId81dea63b8ce3e51adf66cf7b9982a248,规定eqId1510639120a1883e66f13794a9df9179,例如eqId5ab75e89a0930b7fd7d2b07edbe8615e,eqId6e53f66a1b8bcb2e30792a52d597808a.则eqId0d56164859cfe93d142fa6a4125ad362_______

三、解答题

17.解下列方程:

(1)eqIdbb7b57663930839b702d120b028ec9f1

(2)eqId2cf0b7cae5e7f26929815c21109b92d1

18.先化简,再求值:eqIdfde411d932226cfaffff93b707b22b45,其中eqIdf36a91b78ea833d5b09c11366324a845.page number 1

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19.上课时老师在黑板上书写了一个分式的正确化简结果,随后用手掌盖住了一部分,形式如下:

@@@6456c549-f415-4eaa-82c0-5df19e2f91d2eqId65c9200b4416471cc73b44c50902d968

(1)聪明的你请求出盖住部分的代数式;

(2)当eqId707ea658f3a9359f5740d5aab48f7948,eqIdd053b14c8588eee2acbbe44fc37a6886等于何值时,原分式的值为5?

20.已知代数式eqId60ddf3203b8710dfdf7a3e4ee3e5de6a.

(1)化简M;

(2)判断M的值是否可能为eqId274a9dc37509f01c2606fb3086a46f4f,请说明理由.

21.在化简eqIdc8e347fb173ca4d7294a3f54ef0a0f2d时,两位同学分别写出如下第一步运算步骤:

小深:原式eqId360fde0d29a9cad4f7339f0915c8d07b

小圳:原式eqIda275d69e696ef34bf4702b4b219cd068

(1)小深解法第一步的依据是______,小圳解法第一步的依据是______.

A.等式的基本性质        B.分式的基本性质        C.乘法结合律        D.乘法分配律

(2)请你从小深和小圳的两种解法中选择一种解法,接着写出完整的解答过程,并从“3,eqId81fb134b2b48acc99213fff6ccfee65f,1,eqIdacbc6a613224461ade69362d46550474”中选一个合适的数作为x的值,代入求该分式的值.

22.eqIdbfbacb056c2f94379c5f552638d9058d年春节,智能健康手表成为热门“孝心年货”,其中eqId5963abe8f421bd99a2aaa94831a951e9、eqId7f9e8449aad35c5d840a3395ea86df6d两款手表深受市民喜爱.某商店专营该两款手表,已知eqId7f9e8449aad35c5d840a3395ea86df6d款手表的进价比eqId5963abe8f421bd99a2aaa94831a951e9款手表每块多eqId4b72ac611ae66b86761e080761d9aabc元.该商店用eqIdb8cf7857e905e95b52a7f8cb6632c9a8元购进eqId5963abe8f421bd99a2aaa94831a951e9款手表的数量,与用eqId9de64858d26b22085a3d3ddc19112675元购进eqId7f9e8449aad35c5d840a3395ea86df6d款手表的数量相等.

(1)求eqId5963abe8f421bd99a2aaa94831a951e9款、eqId7f9e8449aad35c5d840a3395ea86df6d款手表每块的进价分别为多少元?

(2)该商店计划购进这两款手表(两种都要购进)共eqId0e6f1af4b44b2e97e8f319bab4ae9010块,且进货总费用不超过eqIde5aefee89f26b7875bde0c35965e14de元.已知每块eqId5963abe8f421bd99a2aaa94831a951e9款手表利润eqIde1f959e5f8d89390f0f136f6acc9f6fb元,每块eqId7f9e8449aad35c5d840a3395ea86df6d款手表利润eqId0efba7147f5b9ced8bc4a72f0a9fb8af元.求全部售出后可获得的最大总利润.

23.阅读理解题.

我们定义:如果两个分式eqId5963abe8f421bd99a2aaa94831a951e9与eqId7f9e8449aad35c5d840a3395ea86df6d的差为常数,且这个常数为正数,则称eqId5963abe8f421bd99a2aaa94831a951e9是eqId7f9e8449aad35c5d840a3395ea86df6d的“和谐式”,这个常数称为eqId5963abe8f421bd99a2aaa94831a951e9关于eqId7f9e8449aad35c5d840a3395ea86df6d的“和谐值”.

例:分式eqId40a503ed6f0cd67c76709490aafe7bb5,则eqId5963abe8f421bd99a2aaa94831a951e9是eqId7f9e8449aad35c5d840a3395ea86df6d的“和谐式”,eqId5963abe8f421bd99a2aaa94831a951e9page number 2

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