11.5因式分解 同步练习 -2026-2027学年华师大版八年级数学上册

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11.5因式分解

学校:___________姓名:___________班级:___________考号:___________

一、单选题

1.下列从左到右的变形属于因式分解的是(     )

A.eqIdbfefda7a6cd8ac8dd4de69f1930f30de     B.eqId705edb69682bb8bc91d0b05931a399c6    C.eqIde54809eb946a9742153cab8e4f3d224e   D.eqId7a235cc46f99df5163a4e24f70f90435

2.把多项式eqId832486a8aee3b24a1b08728fa4352525分解因式,应提的公因式是(     )

A.eqId4862a4aad3abc73999aaee9aa304bd58 B.eqIde70eab7a89a5ef3af00a64bfacdad12b C.eqId6b53f61750c2b0b552ddb6d22f12c1dd D.eqIda08a6b56c8980ed2e12adc6a43482097

3.下列因式分解正确的是(     )

A.eqId63595c41fce709bb2d7575a12a2c5081   B.eqId40f5c1084c65b90b84f54297ddef437d  C.eqId3b9fcc1f32877448ad35750fab78c772  D.eqIdc421cfd8c25fe7f831bfa75e6864c78a

4.下列多项式中,不能用公式法因式分解的是(     )

A.eqId55bdeb4b764c8a723d7984337cfd0583       B.eqId73e68a435e4f4c481f3c556dcce6735b       C.eqIde6e08d5bc6be824b0990bac662c661c2      D.eqId296a2c52717297c597321e026c56d022

5.若多项式eqIddae780b566ef9725bed0d7d922d88e77可以用平方差公式进行因式分解,则eqId5963abe8f421bd99a2aaa94831a951e9表示的单项式可以是(     )

A.eqId63b2a6242fecc960fba0c0dd3368f534 B.eqId3e0787af8620f22746ef837f423e97dd C.eqId2c8dc0c98d8a764735afe48c65bd6dd4 D.eqId457c50ab68909712412a30cd19d7c01e

6.小强是一位密码编译爱好者,在他的密码手册中,有这样一条信息:eqIdc9f2416d1f75a45a314331146550832e,eqId059d6df06a5b85848dc4fa33327f8e07,eqIdb88584cf1df43e28d03592c7998b1653,eqId745c100afb56c39c7706fc59aacac0ba分别表示河、爱、南、我.现分解因式eqIde9435e3f058ef4cc6abf63f607524016,结果呈现的密码信息可能是(     )

A.我爱河 B.我河南 C.我爱河南 D.爱河南

7.有两个正方形eqIdff4489d9b83072184c0e1d6b09be50ca、eqIdea1e8babee63bfc889ae5a34632284bc,将eqIdea1e8babee63bfc889ae5a34632284bc放在eqIdff4489d9b83072184c0e1d6b09be50ca的内部得图1,将eqIdff4489d9b83072184c0e1d6b09be50ca、eqIdea1e8babee63bfc889ae5a34632284bc并列放置后得图2,如果图1和图2中阴影部分的面积分别为21和6,则正方形eqIdff4489d9b83072184c0e1d6b09be50ca、eqIdea1e8babee63bfc889ae5a34632284bc的面积之和是(     )

@@@75a48df6-27ff-4076-9e20-bdd655c0decb

A.27 B.28 C.29 D.30

8.已知eqId661177eb83e8726d5ca38e90cb69df61则按此规律推算eqIdf18448312270d0b143dbcb28603de83d的结果一定能(     )

A.被12整除 B.被13整除 C.被14整除 D.被15整除

9.若a、b、c为三角形的三条边长,且eqId1052dd4c18f7e0e943c8edfc115eaef1,则m的取值范围为(     )

A.eqId4f4c78214e43a8b93f2a57072033cbcf B.eqId58b140e221ddf537b8964fff8557cca0 C.eqIdbe466586da8810ccfd811c59a747adb8 D.eqId58b140e221ddf537b8964fff8557cca0或eqId4f4c78214e43a8b93f2a57072033cbcf

10.我们约定:若一个正整数能表示成两个偶数的平方差,那么我们称这个正整数为“美丽数”,如eqId6d97db6e217a9401024dd82ced228854,所以4就是一个“美丽数”.下列数字中,不是“美丽数”的是(     )

A.36 B.28 C.40 D.48page number 0


11.已知eqId8792eaab0b6464e5d07436c64aa751a2,eqIdd26bd6e5455614942759d5354b0c9188,可得eqIdd38cfacb377d5b320df5ecd63b6fa6a9(     )

A.eqIdea1834490aacbee800ed5721312f4be1 B.eqId4b837fd9c52f60bfb3b6852733abc790 C.eqId58b184c94e38f1e5dbe750b2168c2a37 D.eqId8da45c443af7994a26ffa9d8894e7262

12.将下列多项式因式分解,结果中不含有因式eqIda29bb641256754ef10d9012e5fe54a8c的是(     )

A.eqId1ad82fa37dff30a4c54465ff7ef9f7ec       B.eqId6f09b3b88c1d041fe45c218d0ac3b3a4        C.eqId3b7baab4195f5921a4a5e7aaa217e210     D.eqId42e65d25f07f625c27bc92e49815d02d

二、填空题

13.多项式eqIdc1d11d25aef97d915454953cdd487420有一个因式是eqId73b5a9de8a72dda70e3eefeb0a408ffd,则eqIdf0a532e15e232cb4b99a8d4d07c89575的值为__________.

14.若eqId81dea63b8ce3e51adf66cf7b9982a248,eqIdd053b14c8588eee2acbbe44fc37a6886满足eqId9f670f3468ecf566c6b0ca622be0b2e0,则eqId745c100afb56c39c7706fc59aacac0ba的值为_______.

15.若eqId6902f776200a789cdc3ae6ee589e1778,eqId4d641f8c99979acc18997c45d410389d,则eqIda01d7eab3cda0aeadcae94eff8ebe370__________.

16.分解因式:eqIdfa33dd25f60f440062af85adf5b3dab9___________.

三、解答题

17.分解因式:

(1)eqIdb4d66e9fde3f6a5c6ea6c72b8c0ec722       (2)eqId26eea9fba92958502a5642aac3f3f720         (3)eqId5282aa14cec1a7f5b72d382630ddfc91          (4)eqIdc3d54906898bc507acb43dc032b6f58b

 

 

 

 

 

 

 

18.因式分解:

(1)eqId04032b35a698e4c389a44eda265a5924.    (2)eqIde4fb2fcae65b311a90effb93b3100a91       (3)eqIdb188714efbcf26f425e3be0fe7268f46      (4)eqIdcdc6a9c432bba9c6727fadc6cb7212c1

 

 

 

 

 

 

 

 

19.用简便方法计算:

(1)eqId75719fd671bf7e5124f58aba6d35d5f1;                           (2)eqId4a5d799dae391b7f20a2249b41ff1ed3

 

 

 

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20.先阅读材料,再回答问题.

分解因式:eqId8b123ec2e0444de30264ea892d060de8.

解:将eqIdc9f2416d1f75a45a314331146550832e看成整体,令eqIdc5edfe9d5a07a88a2c7eb7e870c61484,则原式eqId39569c0715a97dfaa61709af110f10cc,再将eqIdc5edfe9d5a07a88a2c7eb7e870c61484还原,得到:原式eqId74751b2fa64e95f72753b40105f068f2

上述解题过程中用到了整体思想,它是数学中常用的一种思想.请你用整体思想解决下列问题:

(1)因式分解:eqIde583f445eecbd5efaf7392579abd0363_______.

(2)因式分解:eqId99a07fe3089f7c7e3a8d3ada6fefb1b5;

(3)若n为正整数,则eqId45d08d69b15f74a76d1cdff237b8db6b的值为某一个正整数的平方.请说明理由.

 

 

 

 

 

 

 

 

 

 

21.阅读材料:

eqId2bff73a8ac8ad681a26ed1c2ed457d5feqIdad33ecf186115f11e2811dd4f36e2027

上面的方法称为多项式的配方法,运用多项式的配方法及平方差公式能对一些多项式进行因式分解.根据以上材料,解答下列问题:

(1)因式分解:eqId527ed5ec08bfac2d403309c2ee8256b6;

(2)求多项式eqIdab74300093d8368da472e23150bb7309的最小值;

(3)已知eqId0a6936d370d6a238a608ca56f87198de、eqId2c94bb12cee76221e13f9ef955b0aab1、eqId071a7e733d466949ac935b4b8ee8d183是eqId15c0dbe3c080c4c4636c64803e5c1f76的三边长,且满足eqIda7e1980c58e474d447030f7c69420b6b,求eqId15c0dbe3c080c4c4636c64803e5c1f76的周长.

 

 

 

 

 

 

 

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