2.2一元二次方程的解法(第2课时)同步练习2026-2027学年苏科版数学九年级上册

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2.2一元二次方程的解法(第2课时)同步练习

 

一、单选题

1.用配方法解一元二次方程eqIdb4e4f5a72c45fd8e9d6d2726a09bdf2e时,第一步变形后应是(    )

A.eqIdd9a28b2028adbf0518006cdd94630784 B.eqIdce386f49f71217859bd3f69e89efb2b6 C.eqIdd37917f70914764af1189b5c881ed7ce D.eqId860aeb601bf428551e9ded36eb83565e

2.用配方法解一元二次方程eqIdb45f3e82b25339b98d64f6c52084c530,此方程可化为(     )

A.eqIdda87e6d0122b30a7f611816e44029f03 B.eqId719aa8007c1e57e702316fcfdbddcd53 C.eqId7fc191eec8eac0006a9e26d46ebb18b9 D.eqId66c9a662435c3fd6eebddecc35024a75

3.将式子eqId14ce239e45dc944ad14c683942cb20f6化为eqIdd8676e2a5a165989ee7edb5a72bc9fbd的形式,其结果为(    )

A.eqId8c2c88714f4e2bf32334755c64656287 B.eqIdfac029972213a7ae82c90d03562f1a18 C.eqId86c1211c6bf414dc2528d8c4f6f403e8 D.eqId371cef451a9c8e0004cd5dcf2b1dbfb5

4.若一元二次方程eqIddffe4aaf0a725f8b811ffb25de957e42可转化为eqId61d5554678b16208bb16c6013c37aa06的形式,则eqId20d6fc9b90f370fbb27552876b650f8f的值为(   )

A.12 B.9 C.6 D.0

5.将一个关于x的一元二次方程配方为eqId798ce6272ce93b09deb17b2fbea4bcf4,若eqIddedaada32b74e7d783afa1412b754488是该方程的两个根,则p的值是(    )

A.2 B.4 C.eqIda7ffe8515ff6183c1c7775dc6f94bdb8 D.3

6.方程eqId0f91dd17b6c73a15da79246d91ff2cce的解是(    )

A.eqId56147396e10233ae9e7f1e73be7d7614 B.eqId3487ea294f41e569ae217c340ef4eb5a

C.eqIde1eb23233ce791c5da93202bf5f6c73d D.eqIdf695bd3ae5731bcd9ffaeade4b70a269

7.将方程eqIda8fefc5b9154ba6c5b8f966a995290dc用配方法化为eqId8afad707dbaa7e9e41aa860d07607128的形式,则eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1的值为(     )

A.eqId8e258ab9e600435b37465092243d99f6,eqIdfcdb7a488910743dc5c63afb394b87e2 B.eqId8e258ab9e600435b37465092243d99f6,eqId71a034ac7d59436cebc4759fd850dfb5 C.eqIdf22a4a0dd7307a1323d25331e60782d8,eqId71a034ac7d59436cebc4759fd850dfb5 D.eqIdf22a4a0dd7307a1323d25331e60782d8,eqIdfcdb7a488910743dc5c63afb394b87e2

8.一元二次方程eqId5fa47a89676d117546f659b6b2ab0fd2可以转化为两个一元一次方程,若其中一个一元一次方程为eqId69bcf382e1ba584356e4f9aa38aa9a82,则另一个一元一次方程为(   )

A.eqId4980f8fff7d81510e505ef4f62f48223 B.eqId84f95eebded78b103d8cb77ae92086fa C.eqIdde5c03a1c4ae323bddbb1ca67970288d D.eqId55a5eefe6af8d698d1576db4b034064e

 

二、填空题

9.eqId5290195269a240c59e411d276efca87e___________eqId8c40c5c9abb3ffe621d511d481e93905___________eqIdf0baf29da761dc5ef6386a120bc0a27e.page number 0

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10.关于x的一元二次方程eqIde03a3057e0917185548f457318df0d71的解为______.

11.若将方程eqId3897e8efc65b7bbb7a9dfdd9621aebd1化为eqId5bac5c27fdc082888fbea849bc26c0c6,则m=___________.

12.用配方法将一元二次方程eqId2e20a1bffcffb243c0e28d82c3e9768b化为eqIda70b2f52b2de7b67b77f631bac0cdbe0的形式为______.

 

三、解答题

13.解下列一元二次方程:

(1)eqId62b40b0718700c490e6c0c9ceb4d3fe3;

(2)eqIdb981c9fa52b21c0de5f46dbc7d101fe3;

(3)eqId4b0b663c96a45e8031dde30026ad204a;

(4)eqId240956841c0bc6311510c770eb9182ac.

 

 

14.若eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1,eqId071a7e733d466949ac935b4b8ee8d183为eqId15c0dbe3c080c4c4636c64803e5c1f76的三边长,且eqIdb92fd3514b511793a1249a3b51e1357e,试判断eqId15c0dbe3c080c4c4636c64803e5c1f76的形状,并说明理由.

 

 

 

15.解方程eqId60eacc0cce87dae390481f5e46f330cb时,我们可以将eqIdb1b8d3f7a166ff5db92d9ee0014a960d看成一个整体,设eqId714e5445c29d9d1ff3adfc1a784bbb53,则原方程可化为eqId6d508887ede441f4d498565ceeb8682a,解得eqIdcda6ab0f37a6a39baffa6c98dc11dc16,eqIdaeb68f02e333c48c66a2656d5115cf46.当eqIdc50866229ec5a3640fb250f9bd2192b3,即eqId36d6231bf008a53b48ce5af1b5401607,解得eqId707ea658f3a9359f5740d5aab48f7948;当eqIdde72a5190834f5dbe895596656c038b3,即eqId01095b657338cd37efd644a1cb6b7177,解得eqIdda322ac8867e8a47c6588601078abf18.所以原方程的解eqIdf0736457346c11dd6f458418a4f747ff,eqId29fd610958e41038390748e2e082aea1.

(1)请你利用这种方法解方程:eqId982b6fcebb22419d1504d0fdc313a3f3;

(2)已知eqId15c0dbe3c080c4c4636c64803e5c1f76三条边的长度分别为eqId0a6936d370d6a238a608ca56f87198de,eqId2c94bb12cee76221e13f9ef955b0aab1,eqId071a7e733d466949ac935b4b8ee8d183,若满足eqId024b1acef9f30bf78dde1beb191dddd6,且eqId6de1d395e6c48c0676a1488a299479d9,请判断eqId15c0dbe3c080c4c4636c64803e5c1f76的形状,并说明理由.

 

 

 

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16.阅读材料:

为解方程eqIdfa087beea13851d807c073825e29e787,我们可以将eqId9e0d30c29fd988b1126bbd88b5ebf29b视为一个整体,然后设eqIde8da7c06be71acd821a73fadba4d1c35,则eqId3cf83a1e1a93986c879bdf9052b162ba,原方程可化为eqId6d508887ede441f4d498565ceeb8682a,eqId90ee9f77baae294578f7c1ecedd4e52c①

解得eqId2913f8d730167b2bc9bbbbf809de4b1a,eqId2fb664c266129cfa83f5d2c6ba3d7a38.

当eqIdde72a5190834f5dbe895596656c038b3时,eqId35f227df33998c627b586997a32d8f62,eqIdcb571d0bafa48edd17eb9c0fd71fe897,eqId8ec73e76ff893b7b25a2f954757dbf76.

当eqIdc50866229ec5a3640fb250f9bd2192b3时,eqIdfbce8eb03871d4563436ddd3c2eb1808,eqId2607f60f096a69ad082748b8599f7e39,eqIdf76af3c07409e8d7907bbe84156d8e4c.

eqId2de0d10ef8b748d4531250c37c5d3f9e原方程的解为eqIde370d4caeeab628e65b82b0104114aad,eqIdd8e9eb18f22ed5b7927a0d0bfa988bb0,eqId42cb565d4ee7a36e30865423e8524558,eqIda1c6765d414b74c91d0ad0f06db034a3.

以上方法就叫换元法,体现了转化的思想.

请仿照材料解决问题:

(1)解方程:eqId8f4232581018b353f5a78469f3f46484.

(2)若四个连续正整数的积为eqIdcfd121f4aeca8a78a320fca1fd1c4d07,求出这四个连续的正整数.

 

 

 

 

17.配方法是数学中重要的一种思想方法.它是指将一个式子的某一部分通过恒等变形化为完全平方式或几个完全平方式的和的方法.这种方法常被用到代数式的变形中,并结合非负数的意义来解决一些问题.先阅读理解下面的例题,再按要求解答问题.

例:求代数式eqIdfc67e38d9230b7ce4f46a8c819d0b85f的最小值.

解:eqId6d51166e4ec27e97a7b2897cc2718ab3.

因为eqIdccf82104cabc9da441bdd0edc99eddba,所以eqId21950ea238e48e30f3036a40364f8b4f,所以eqIdfc67e38d9230b7ce4f46a8c819d0b85f的最小值是eqId61128ab996360a038e6e64d82fcba004.

(1)代数式eqId00e592d76fd3ef30e964393a1eff2624的最小值为 .

(2)关于eqId81dea63b8ce3e51adf66cf7b9982a248的二次多项式eqId98d257a86e2f146ebccc6ce517c64c4d(eqId0a6936d370d6a238a608ca56f87198de为常数)有最小值为eqIdd3aab28a20b5bf47040aaec03b1eb550,求常数eqId0a6936d370d6a238a608ca56f87198de的值.

(3)已知实数eqId81dea63b8ce3e51adf66cf7b9982a248,eqIdd053b14c8588eee2acbbe44fc37a6886满足eqId61c178b4eaed89fa32d0eb7e55c63735,求eqId61e3da18d31497e7c2ce5217d139f80e的最大值.

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