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第二章因式分解练习题一.填空题:1.1218323xyxy的公因式是___________2.分解因式:2183xx__________3.若AxyByx353,,则AABB222_________4.若xxt26是完全平方式,则t=________5.因式分解:944222abbcc_________6.分解因式:acabcabc32244_________7.若||xxxyy214022,则x=_______,y=________8.若ab9998,,则aabbab22255_________9.计算12798012501254798....________10.运用平方差公式分解:a2-_______=(a+7)(a-_____)11.完全平方式49222xy()12.若a.b.c,这三个数中有两个数相等,则abcbcacab222()()()_____13.若abab514,,则aababb3223__________二.选择题:14.下列各式从左到右的变形为分解因式的是()A.18363232xyxyB.()()mmmm2362C.xxxxx289338()()D.mmmm2623()()15.多项式36322xyxyxy提公因式3xy后另一个多项式为()A.xy2B.xy21C.xy2D.xy2116.下列多项式中不含有因式()x1的是()A.2313xxB.xx245C.xx287D.xx2617.下列各式进行分解因式错误的是()A.96322()()()xyxyxyB.41292222()()()abaabaabC.()()()()()ababacacbc2222D.()()()mnmnmn2221118.()()aaamm1的值是()A.1B.-1C.0D.()11m19.把3154521aaannn分解因式是()A.35152aaan()B.351521aaan()C.12D.35151aaan()20.若n为任意整数,()nn1122的值总可以被k整除,则k等于()A.11B.22C.11或22D.11的倍数21.下列等式中一定正确的是()A.()()abbannB.()()abbannC.()()baabnnD.()()ababnn22.多项式8102233222mnmnmn被222mn除,所得的商为()A.451nmB.451nmC.451nmD.45nm三.解答题(共61分)23.把下列各式分解因式:(1)mmnnm2224()()(2)xxyy22444(3)()()343272222xxxx(4)xxx3214(5)xxxxxxx()()()11113224.计算:(1)2222998101100(2)2004220042002200420042005323225.已知mn3,mn23,求mnmnmn3223的值。(10分)【试题答案】一.填空题1.62xy2.233xxx()()3.4322()xy4.95.()()3232abcabc6.acab()227.2,48.-49.110.49,711.12xy,2x-3y12.013.265二.选择题14.D15.D16.D17.D18.C19.A20.A21.A22.C三.解答题23.(1)解:原式mmnmn2224()()mmnmnmnmmnmm222222222222()()()()()()()(2)解:原式()xxyy22444()()()xyxyxy22222222(3)解:原式[()()][()()]34327343272222xxxxxxxx()()55431022xxxx(4)解:原式xxxxx()()221412(5)解:原式()[()()]xxxxxx11112(){()[()]}()[()()]()()()xxxxxxxxxxx1111111111222424.计算(1)解:原式222222212212221498981001009810098100()()(2)解:设a=2004则原式aaaaaaaaaaaaaa323222111211121()()()()()()()()将a=2004代入得原式2002200525.解:mnmnmn3223mnmmnnmnmnmnmnmnmnmn()()[()]22222233将mnmn323,代入得原式23332323921432[][]26.(1)解:原式xxyyxzyzz222694124()()()xyzxyzxyz343432222(2)解:原式()()aaaa222510524120()()[()][()]aaaaaaaa2222251059651656()()()()()()aaaaaaaa225166116516
本文标题:八年下第二章分解因式01
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