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因式分解练习题加答案-200道

因式分解3a3b2c-6a2b2c2+9ab2c3=3ab^2 c(a^2-2ac+3c^2)3.因式分解xy+6-2x-3y=(x-3)(y-2)4.因式分解x2(x-y)+y2(y-x)=(x+y)(x-y)^25.因式分解2x2-(a-2b)x-ab=(2x-a)(x+b)6.因式分解a4-9a2b2=a^2(a+3b)(a-3b)7.若已知x3+3x2-4含有x-1的因式,试分解x3+3x2-4=(x-1)(x+2)^28.因式分解ab(x2-y2)+xy(a2-b2)=(ay+bx)(ax-by)9.因式分解(x+y)(a-b-c)+(x-y)(b+c-a)=2y(a-b-c)10.因式分解a2-a-b2-b=(a+b)(a-b-1)11.因式分解(3a-b)2-4(3a-b)(a+3b)+4(a+3b)2=[3a-b-2(a+3b)]^2=(a-7b)^212.因式分解(a+3)2-6(a+3)=(a+3)(a-3)13.因式分解(x+1)2(x+2)-(x+1)(x+2)2=-(x+1)(x+2)abc+ab-4a=a(bc+b-4)(2)16x2-81=(4x+9)(4x-9)(3)9x2-30x+25=(3x-5)^2(4)x2-7x-30=(x-10)(x+3)35.因式分解x2-25=(x+5)(x-5)36.因式分解x2-20x+100=(x-10)^237.因式分解x2+4x+3=(x+1)(x+3)38.因式分解4x2-12x+5=(2x-1)(2x-5)39.因式分解下列各式:(1)3ax2-6ax=3ax(x-2)(2)x(x+2)-x=x(x+1)(3)x2-4x-ax+4a=(x-4)(x-a)(4)25x2-49=(5x-9)(5x+9)(5)36x2-60x+25=(6x-5)^2(6)4x2+12x+9=(2x+3)^2(7)x2-9x+18=(x-3)(x-6)(8)2x2-5x-3=(x-3)(2x+1)(9)12x2-50x+8=2(6x-1)(x-4)40.因式分解(x+2)(x-3)+(x+2)(x+4)=(x+2)(2x-1)41.因式分解2ax2-3x+2ax-3=(x+1)(2ax-3)42.因式分解9x2-66x+121=(3x-11)^243.因式分解8-2x2=2(2+x)(2-x)44.因式分解x2-x+14 =整数内无法分解45.因式分解9x2-30x+25=(3x-5)^246.因式分解-20x2+9x+20=(-4x+5)(5x+4)47.因式分解12x2-29x+15=(4x-3)(3x-5)48.因式分解36x2+39x+9=3(3x+1)(4x+3)49.因式分解21x2-31x-22=(21x+11)(x-2)50.因式分解9x4-35x2-4=(9x^2+1)(x+2)(x-2)51.因式分解(2x+1)(x+1)+(2x+1)(x-3)=2(x-1)(2x+1)52.因式分解2ax2-3x+2ax-3=(x+1)(2ax-3)53.因式分解x(y+2)-x-y-1=(x-1)(y+1)54.因式分解(x2-3x)+(x-3)2=(x-3)(2x-3)55.因式分解9x2-66x+121=(3x-11)^256.因式分解8-2x2=2(2-x)(2+x)57.因式分解x4-1=(x-1)(x+1)(x^2+1)58.因式分解x2+4x-xy-2y+4=(x+2)(x-y+2)59.因式分解4x2-12x+5=(2x-1)(2x-5)60.因式分解21x2-31x-22=(21x+11)(x-2)61.因式分解4x2+4xy+y2-4x-2y-3=(2x+y-3)(2x+y+1)62.因式分解9x5-35x3-4x=x(9x^2+1)(x+2)(x-2)63.因式分解下列各式:(1)3x2-6x=3x(x-2)(2)49x2-25=(7x+5)(7x-5)(3)6x2-13x+5=(2x-1)(3x-5)(4)x2+2-3x=(x-1)(x-2)(5)12x2-23x-24=(3x-8)(4x+3)(6)(x+6)(x-6)-(x-6)=(x-6)(x+5)(7)3(x+2)(x-5)-(x+2)(x-3)=2(x-6)(x+2)(8)9x2+42x+49=(3x+7)^2 。

1.若(2x)n−81 = (4x2+9)(2x+3)(2x−3),那么n的值是( B)A.2 B.4 C.6 D.82.若9x2−12xy+m是两数和的平方式,那么m的值是( B)A.2y2 B.4y 2 C.±4y2 D.±16y23.把多项式a4− 2a2b2+b4因式分解的结果为( D)A.a2(a2−2b2)+b4 B.(a2−b2)2C.(a−b)4 D.(a+b)2(a−b)24.把(a+b)2−4(a2−b2)+4(a−b)2分解因式为( C )A.( 3a−b)2 B.(3b+a)2C.(3b−a)2 D.( 3a+b)26.已知x,y为任意有理数,记M = x2+y2,N = 2xy,则M与N的大小关系为( B) A.M>N B.M≥N C.M≤N D.不能确定7.对于任何整数m,多项式( 4m+5)2−9都能( A )A.被8整除B.被m整除C.被(m−1)整除D.被(2n−1)整除9.下列变形中,是正确的因式分解的是( D )A.0.09m2− n2 = ( 0.03m+ n )( 0.03m−n)B.x2−10 = x2−9−1 = (x+3)(x−3)−1C.x4−x2 = (x2+x)(x2−x)D.(x+a)2−(x−a)2 = 4ax10.多项式(x+y−z)(x−y+z)−(y+z−x)(z−x−y)的公因式是( A)A.x+y−z B.x−y+z C.y+z−x D.不存在11.已知x为任意有理数,则多项式x−1−x2的值( )A.一定为负数B.不可能为正数C.一定为正数D.可能为正数或负数或零二、解答题:分解因式:(1)(ab+b)2−(a+b)2(2)(a2−x2)2−4ax(x−a)2(3)7xn+1−14xn+7xn−1(n为不小于1的整数)答案:一、选择题:1.B 说明:右边进行整式乘法后得16x4−81 = (2x)4−81,所以n应为4,答案为B.2.B 说明:因为9x2−12xy+m是两数和的平方式,所以可设9x2−12xy+m = (ax+by)2,则有9x2−12xy+m = a2x2+2abxy+b2y2,即a2 = 9,2ab = −12,b2y2 = m;得到a = 3,b = −2;或a = −3,b = 2;此时b2 = 4,因此,m = b2y2 = 4y2,答案为B.3.D 说明:先运用完全平方公式,a4− 2a2b2+b4 = (a2−b2)2,再运用两数和的平方公式,两数分别是a2、−b2,则有(a2−b2)2 = (a+b)2(a−b)2,在这里,注意因式分解要分解到不能分解为止;答案为D.4.C 说明:(a+b)2−4(a2−b2)+4(a−b)2 = (a+b)2−2(a+b)[2(a−b)]+[2(a−b)]2 = [a+b−2(a−b)]2 = (3b−a)2;所以答案为C.6.B 说明:因为M−N = x2+y2−2xy = (x−y)2≥0,所以M≥N.7.A 说明:( 4m+5)2−9 = ( 4m+5+3)( 4m+5−3) = ( 4m+8)( 4m+2) = 8(m+2)( 2m+1).9.D 说明:选项A,0.09 = 0.32,则0.09m2− n2 = ( 0.3m+n)( 0.3m−n),所以A 错;选项B的右边不是乘积的形式;选项C右边(x2+x)(x2−x)可继续分解为x2(x+1)(x−1);所以答案为D.10.A 说明:本题的关键是符号的变化:z−x−y = −(x+y−z),而x−y+z≠y+z−x,同时x−y+z≠−(y+z−x),所以公因式为x+y−z.11.B 说明:x−1−x2 = −(1−x+x2) = −(1−x)2≤0,即多项式x−1−x2的值为非正数,正确答案应该是B.二、解答题:(1) 答案:a(b−1)(ab+2b+a)说明:(ab+b)2−(a+b)2 = (ab+b+a+b)(ab+b−a−b) = (ab+2b+a)(ab−a) = a(b−1)(ab+2b+a).(2) 答案:(x−a)4说明:(a2−x2)2−4ax(x−a)2= [(a+x)(a−x)]2−4ax(x−a)2= (a+x)2(a−x)2−4ax(x−a)2= (x−a)2[(a+x)2−4ax]= (x−a)2(a2+2ax+x2−4ax)= (x−a)2(x−a)2 = (x−a)4.(3) 答案:7xn−1(x−1)2说明:原式= 7xn−1 •x2−7xn−1 •2x+7xn−1 = 7xn−1(x2−2x+1) = 7xn−1(x−1)2.因式分解之十字相乘法专项练习题(1)a2-7a+6;(2)8x2+6x-35;(3)18x2-21x+5;(4) 20-9y-20y2;(5)2x2+3x+1;(6)2y2+y-6;(7)6x2-13x+6;(8)3a2-7a-6;(9)6x2-11x+3;(10)4m2+8m+3;(11)10x2-21x+2;(12)8m2-22m+15;(13)4n2+4n-15;(14)6a2+a-35;(15)5x2-8x-13;(16)4x2+15x+9;(17)15x2+x-2;(18)6y2+19y+10;(19) 2(a+b) 2+(a+b)(a-b)-6(a-b) 2;(20)7(x-1) 2+4(x-1)-20;(1)(a-6)(a-1),(2)(2x+5)(4x-7)(3)(3x-1)(6x-5),(4)-(4y-5)(5y+4)(5)(x+1)(2x+1),(6)(y+2)(2y-3)(7)(2x-3)(3x-2),(8)(a-3)(3a+2)(9)(2x-3)(3x-1),(10)(2m+1)(2m+3)(11)(x-2)(10x-1),(12)(2m-3)(4m-5)(13)(2n+5)(2n-3),(14)(2a+5)(3a-7)(15)(x+1)(5x-13),(16)(x+3)(4x+3)(17)(3x-1)(5x=2),(18)(2y+5)(3y+2)(19)(3a-b)(5b-a),(20)(x+1)(7x-17)例1 分解因式思路1 因为所以设原式的分解式是然后展开,利用多项式的恒等,求出m, n,的值。

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