第九章因式分解单元测试2025-2026学年苏科版数学八年级下册

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                            第九章  因式分解 单元练习

一、单选题

1.下列各式由左到右的变形中,属于因式分解的是(    )

A.eqId0579f5886cfc0f27afb12d8316f39a35 B.eqIdc3f7b77c4e9fb73ece56d1ad954b85ac

C.eqIda969026394a518490912fa412c4ad9b5 D.eqId286dd8d13a7ff9ee7749e6bdc5542427

2.分解因式eqId627a18b125f6a0c1b9fb360680b27786结果正确的是(    )

A.eqIdd2a187c83d454246110bb388a348969d B.eqIde4ae066c73270b038400eaf6e606574f C.eqIdbda21cca551103770104c861e868b077 D.eqIddd8279caee1b2fb03c7b4e590cc03a53

3.下列分解因式的变形中,正确的是(    )

A.xy(x﹣y)﹣x(y﹣x)=﹣x(y﹣x)(y+1)

B.6(a+b)2﹣2(a+b)=(2a+b)(3a+b﹣1)

C.3(n﹣m)2+2(m﹣n)=(n﹣m)(3n﹣3m+2)

D.3a(a+b)2﹣(a+b)=(a+b)2(2a+b)

4.多项式eqIdce599e4ee8d1b450fc18ca98bd05bb1d可因式分解成eqId10d37c397b01be34059a8251786ecf4a,其中eqId0a6936d370d6a238a608ca56f87198de、eqId2c94bb12cee76221e13f9ef955b0aab1、eqId071a7e733d466949ac935b4b8ee8d183均为整数,求eqId74094c51b5ee19992e12e98bfab66d77的值为(  )

A.eqId52c339de3145fd52e82a1887383750df B.eqId5ca7d1107389675d32b56ec097464c14 C.eqId6dbf80cb71406fd14621139149f7ed9e或eqId45e61680ffc6f940bd5b8afda3ae9c59 D.eqId5ca7d1107389675d32b56ec097464c14或eqId45e61680ffc6f940bd5b8afda3ae9c59

5.下列从左边到右边的变形中,是因式分解的是(    )

A.eqId9a26501d15f6852e68b085151f9fde82 B.eqId92ce3f655082539524d967e132237ef8

C.eqIdf1989f93ad728aade6fd98fcf4040f95 D.eqId58bf7c24a279d9c856a1676aeae7d52e

6.下列各式中能用完全平方公式因式分解的是(    )

A.eqIdf8c19360fd839891ef070022491e6e58 B.eqIdb52b2e6b7b6a424575d7b02e9028fb88 C.eqIdf4e1fbe0fb49725cf6d1e689ee8986d6 D.eqIdf034f8aa0402bc5b12f669645d452b73

7.下列因式分解正确的是(   )

A.eqIdd9f94cf00ad88eafbb826c6ff6efa3a9 B.eqId77ada8ac5990065e87752a031721140e

C.eqId30da44f819c40e9de87ee3b830cee3f8 D.eqIdec93195acda42b0e2e6285e8ae448faa

8.对于二次三项式eqIde3588347d3bef0e1b1227291a23f18a2(eqId294f5ba74cdf695fc9a8a8e52f421328为常数),下列结论正确的个数有(   )

①当eqId7aed39f5aca78934fb383402433fe549时,若eqId45994df98fd27e747bbc81af7d1a7b32,则eqIdc9021f0886cadf8d9f6f12d50be1d3d3;

②无论eqId81dea63b8ce3e51adf66cf7b9982a248取任何实数,等式eqIdadf4292f7acb0c2c9dc5b9914aa7fd73都恒成立,则eqId58a2c8fc8f04a10d423dac1cc602959d;

③若eqIdbc317229b1e1f0ae8dd08c0cfc25f87f,eqId8f5b94ba6c0a1f68c3ff47aee641ef85,则eqIdcbe0890043f34b4575bf7bb3a773e32b或eqId274a9dc37509f01c2606fb3086a46f4f;

④满足eqId60e0e515fd8289cb3d6e0b6ebc424c97的整数解eqId82a79a33a83a7ba57a34b5093d1d1d02共有8个.

A.1个 B.2个 C.3个 D.4个

9.若实数eqId81dea63b8ce3e51adf66cf7b9982a248,eqIdd053b14c8588eee2acbbe44fc37a6886满足eqIde00a21c195d7d86c6d0493c77f26aa12,则eqId29ebc856291255f2d4a6c20b982a2442的值为(     )

A.5或eqIdacbc6a613224461ade69362d46550474 B.5 C.1或eqIdd13ce3ebd1112220c639562739f1f9d1 D.1

10.下列各式从左边到右边的变形是因式分解的是(    )

A.eqId19a59bf580963bf4e6887e8a4076dead B.eqId7a02dffa7843aac4e72fdbe6f7857aa2

C.eqId5f867edfac52237487a3f3b41df718e0  D.eqId751bc4c28172b1f276c8ac6e20ab9fa4)

二、填空题

11.将多项式eqId335a46a7a2f76be27b615b5ae83d8221因式分解得________.

12.因式分解eqIdd412dce79f4f5d0c4cc0c4cb86585806______.

13.把多项式eqIdad82a397b587c88ce7390c16d41db75a因式分解,结果为____.

14.将多项式eqId357999b1208f9de0e8319113323bba35因式分解的结果是__________.

15.多项式eqId252a5bd6c6a2bcca5fe98cf2d543e684的最小值为___________.

16.在实数范围内分解因式:eqIdc1123788ad280e735f2fef623eb628a3______.

17.分解因式:eqId246a8f8d20e7b09c5ab1dd5a8d07c13a________.

18.如果两数eqId9b0fffbec1fe851795dfdd448bf0d165满足eqId8d6d275782abf8a0ad35c4b539999a62,那么eqId4b76a763e16272d3a6e9588c7a8c4ec2_______.

三、解答题

19.用简便方法计算:

(1)eqId7a3c1c3a7cd4fd35f56e0e59abef05f7;                  (2)eqId7e6b35f864ef009713e505285b64a02b.

 

 

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20.分解因式

(1)eqIda0ff686c06c526d8f2e51075364fcec9                    (3)eqId71c93c96ad657c385d3c772fc3a919ca                               (4)eqId8e9f31810f4156c38f51b36484801192

 

 

 

 

21.数形结合是解决数学问题的重要思想方法,在学习“因式分解”时,我们可以借助直观、形象的几何模型来求解.下面共有三种卡片:A型卡片是边长为x的正方形,B型卡片是长为y,宽为x的长方形,C型卡片是边长为y的正方形.请用1张A型卡片,2张B型卡片和1张C型卡片拼成一个大正方形,在图2虚线框中画出正方形的示意图,再据此写出一个多项式的因式分解.

@@@1a604dee-80c0-4bf4-977a-3b1090c6e78f

 

 

 

 

 

 

 

 

22.小明的老师设计了一个“神秘数字盒子”,盒子的密码由以下规则生成:将二次多项式eqIdcb02d7da85301dfeebc5833cd98b7d86因式分解为eqId99ad4df95e444375e9a8a4610ea5178d,其中 m,n,p,q均为整数.密码由分解后各因式的系数绝对值按从小到大的顺序排列组成(重复数字只保留一个).例如:若分解结果为eqIddd574eef6783de3dbc2e05773b61b246,则系数绝对值为1,2,3,密码为1,2,3.请解决以下问题:

(1)将多项式eqIde89ab928cc13940d179ca2aa3ceedb7f分解为两个一次因式的乘积.根据分解结果,生成该数字盒子的密码.

(2)若某数字盒子的密码为1,2,3,5,且其对应的多项式为eqId1c79a526a4cc5cfca05869358c28a6eb,请写出该多项式的一种可能的因式分解形式,并写出此时k的值.

 

 

 

 

23.如图,在半径为eqId4aa0df7f1e45f9de29e802c7f19a4f64的圆形钢板上冲去半径为eqId11bc05f41215f9894e11d1df0465751a的四个小圆孔.若eqId6d0828bdee6b9dcc054cc916ac5ae114,eqId32cb20c495b91de77c987d2378b7bad0,请你利用因式分解的方法计算出剩余钢板的面积.(eqId70f5389990c3a0c5373f3bd9fb2454c9取3.14)

@@@106498cc-4d5a-4679-95e4-55105be778d4

 

 

24.已知eqIddf3e31a58d653b6bf558f44d57485928,eqIde6156c3081a530f33b2075e0c7389a58,求eqId92ba636c53c445ee573eff65978135c5的值.

 

 

 

 

 

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25.已知eqIdb6a24198bd04c29321ae5dc5a28fe421为整数,证明代数式eqId6ecb9cf45934f0b2a86b83cb4373e1df的值一定为整数.

 

 

 

 

 

26.先阅读下列材料,然后解决后面的问题.

材料:一个三位数eqIda675d752761f68640cb84085823063bf(百位数为a,十位数为b,个位数为c),若a+c=b,则称这个三位数eqIda675d752761f68640cb84085823063bf为“协和数”,同时规定c=eqId8573eecbc29f522671b3892ec406c50b(k≠0),k称为“协和系数”,如264,因为它的百位上数字2与个位数字4之和等于十位上的数字6,所有264是“协和数”,则“协和数”k=2×4=8.

(1)判断132,123,321这三个数中, 是“协和数”.

(2)对于“协和数”eqIda675d752761f68640cb84085823063bf,求证:“协和数”eqIda675d752761f68640cb84085823063bf能被11整除.

(3)已知有两个十位数相同的“协和数”eqIde99f9a2e743fd0e59a9c6e36f918bb2c,eqId374611d860596eaa4c70fba2bd9fd8b1(eqIde72adb45c60c2f63b46e65ff787302bf>eqId3e88093a749c0d46e0ee931ecfaff925),且eqIdbf209717d3bde602ab96c53d6a43a811,若eqId093c62304e8e248eddf80424369c77eb,用含b的式子表示y.

 

 

 

 

 

 

 

 

 

 

27.阅读:已知eqIdaefb6a113b98d459763c25ddf45c1b69,eqId97c03256f5374101853665aa4c1934e8,求eqIdc925be255ca736a53b24d13ddede1a86的值.

解:∵eqIdaefb6a113b98d459763c25ddf45c1b69,eqId97c03256f5374101853665aa4c1934e8,

∴eqId3afee21404ccf1c129f7971c1fb40e7e.

请你根据上述解题思路解答下面问题:

已知eqIde5178895b1b7670ea686100801de433f,eqIda7a281f6bd5bdcc3afe5555b4c81168b,求eqIdddd5e81ec43ed93fb31b6fedf8324003的值.

 

 

 

 

 

 

 

 

 

 

 

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