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1、 分组法分解因式- 1 分组法分解因式专项练习 30 题(有答案) 1a 2ab+acbc 22x 3x2z4x2y+2xyz+2xy2y2z 3 (x 2+3x+2) (4x2+8x+3)90 4xy+xz+ay+az 5m 24mn+4n21 6x 212ax+a2 7x 3+x2yxy2y3= 8m 2n2+2m2n 9 (1) (2x 23x+1)222x2+33x1; (2)x 4+7x3+14x2+7x+1; (3) (x+y) 3+2xy(1xy)1; (4) (x+3) (x 21) (x+5)20 106x 4+7x336x27x+6 11 (1)2x 5n1yn+4x3n1
2、yn+22xn1yn+4; (2)x 38y3z36xyz; 分组法分解因式- 2 (3)a 2+b2+c22bc+2ca2ab; (4)a 7a5b2+a2b5b7 126x 25xy6y2+2x+23y20 1318a 232b218a+24b 14a 2b2+x2y22(axby) 15a 2b2+2a2b2ab2 16a 22a+14b2 17a 3+ab22a2ba 18x 2y23x3y 19x 2y2xy 20x 3+2x2y4x8y 21254x 2+4xyy2 22a 3(bc)+b3(ca)+c3(ab) 237x 23y+xy21x 242x 33x2+3y22xy2 2
3、5a 28ab+16b2+6a24b+9 分组法分解因式- 3 26m 22mn+n2am+an 27x 22xy+y2+3x3y+2 28(1)a 22ab+b24; (2)x 3x24x+4 29a 2x24+a2y22a2xy 30 (1)x 2+9y2+4z26xy+4xz12yz (2) (a 2+5a+4) (a25a+6)120 分组法分解因式- 4 参考答案参考答案: 1a 2ab+acbc=a(ab)+c(ab)=(ab) (a+c) 22x 3x2z4x2y+2xyz+2xy2y2z=(2x3x2z)(4x2y2xyz)+(2xy2y2z) , =x 2(2xz)2xy(2
4、xz)+y2(2xz)=(2xz) (x22xy+y2)=(2xz) (xy)2 3.原式=(x+1) (x+2) (2x+1) (2x+3)90=(x+1) (2x+3)(x+2) (2x+1)90 =(2x 2+5x+3) (2x2+5x+2)90 令 y=2x 2+5x+2,则原式=y(y+1)90=y2+y90=(y+10) (y9)=(2x2+5x+12) (2x2+5x7) =(2x 2+5x+12) (2x+7) (x1) 4xy+xz+ay+az=xy+xz+ay+az=(xy+xz)+(ay+az)=x(y+z)+a(y+z)=(y+z) (x+a) 5m 24mn+4n21
5、=(m2n)21=(m2n+1) (m2n1) 6x 212ax+a2=(xa)21=(xa+1) (xa1) 7x 3+x2yxy2y3=x3+x2yxy2y3=(x3+x2y)(xy2+y3)=x2(x+y)y2(x+y)=(x+y)2(xy) 8m 2n2+2m2n=(mn) (m+n)+2(mn)=(mn) (m+n+2) 9 (1) (2x 23x+1)222x2+33x1=(2x23x+1)211(2x23x+1)+10=(2x23x+11) (2x23x+110) , =(2x 23x) (2x23x9)=x(2x3) (2x+3) (x3) ; (2)x 4+7x3+14x2+
6、7x+1=x4+4x3+6x2+4x+1+3x3+6x2+3x+2x2=(x+1)22+3x(x+1)2+2x2, =(x+1) 2+2x(x+1)2+x=(x2+4x+1) (x2+3x+1) ; (3) (x+y) 3+2xy(1xy)1=(x+y)31+2xy(1xy)=(x+y1)(x+y)2+x+y+12xy(x+y1) =(x+y1) (x 2+y2+x+y+1) ; (4) (x+3) (x 21) (x+5)20=(x+3) (x+1) (x1) (x+5)20=(x2+4x+3) (x2+4x5)20, =(x 2+4x)22(x2+4x)1520=(x2+4x+5) (x2
7、+4x7) 10、原式=6(x 4+1)+7x(x21)36x2=6(x42x2+1)+2x2+7x(x21)36x2 =6(x 21)2+2x2+7x(x21)36x2=6(x21)2+7x(x21)24x2=2(x21)3x3(x21)+8x =(2x 23x2) (3x2+8x3)=(2x+1) (x2) (3x1) (x+3) 11 (1)原式=2x n1yn(x4n2x2ny2+y4)=2xn1yn(x2n)22x2ny2+(y2)2=2xn1yn(x2ny2)2 =2x n1yn(xny)2(xn+y)2 (2)原式=x 3+(2y)3+(z)33x(2y) (z)=(x2yz)
8、(x2+4y2+z2+2xy+xz2yz) (3)原式=(a 22ab+b2)+(2bc+2ca)+c2=(ab)2+2c(ab)+c2=(ab+c)2 本小题可以稍加变形,直接使用公式,解法如下: 原式=a 2+(b)2+c2+2(b)c+2ca+2a(b)=(ab+c)2 (4)原式=(a 7a5b2)+(a2b5b7)=a5(a2b2)+b5(a2b2)=(a2b2) (a5+b5) =(a+b) (ab) (a+b) (a 4a3b+a2b2ab3+b4)=(a+b)2(ab) (a4a3b+a2b2ab3+b4) 126x 25xy6y2+2x+23y20=6x2x(5y2)(6y2
9、23y+20)=6x2x(5y2)(2y5) (3y4) =(2x3y4) (3x+2y+5) 1318a 232b218a+24b=2(9a216b29a+12b)=2(9a216b2)3(3a4b) =2(3a+4b) (3a4b)3(3a4b)=2(3a4b) (3a+4b3) 14、原式=a 2b2+x2y22ax+2by=(a2+x22ax)(b2+y22by)=(ax)2(by)2 =(a+bxy) (abx+y) 15原式=(a 2b2)+(2a2b2ab2)=(a+b) (ab)+2ab(ab)=(ab) (a+b+2ab) 16原式=(a1) 2(2b)2=(a1+2b) (
10、a12b) 17原式=a(a 2+b22ab1)=a(ab)21=a(ab+1) (ab1) 18原式=(xy) (x+y)3(x+y)=(x+y) (xy3) 19原式=(x 2y2)(x+y)=(x+y) (xy)(x+y)=(x+y) (xy1) 20原式=x 2(x+2y)4(x+2y)=(x24) (x+2y)=(x+2) (x2) (x+2y) 分组法分解因式- 5 21254x 2+4xyy2=25(4x24xy+y2)=52(2xy) 2=(5+2xy) (52x+y) 22a 3(bc)+b3(ca)+c3(ab) =a 3ba3c+b3cb3a+c3ac3b =a 3bb3
11、a(a3cb3c)+c3(ab) =ab(a 2b2)c(a3b3)+c3(ab) =ab(a+b) (ab)c(ab) (a 2+ab+b 2)+c3(ab) =(ab)ab(a+b)c(a 2+ab+b 2)+c3 =(ab)b 2(ac)c(a 2c2)+ab(ac) =(ab) (ac)b 2c(a+c)+ab =(ab) (ac)(b 2c2)+a(bc) =(a+b+c) (ab) (bc) (ca) 237x 23y+xy21x=7x221x+xy3y=7x(x3)+y(x3)=(7x+y) (x3) 24原式=x 2(2x3)+y2(32x)=(2x3) (x2y2)=(2x3
12、) (x+y) (xy) 25原式=(a 28ab+16b2)+(6a24b)+9=(a4b)26(a4b)+9=(a4b3)2 26m 22mn+n2am+an=(m22mn+n2)+(am+an)=(mn)2a(mn)=(mn) (mna) 27x 22xy+y2+3x3y+2=(xy)2+3(xy)+2=(xy1) (xy2) 28 (1)a 22ab+b24=(ab)24=(ab+2) (ab2) ; (2)x 3x24x+4=x34xx2+4=x(x24)(x24)=(x+2) (x2) (x1) 29、 a 2x24+a2y22a2xy=(a2x22a2xy+a2y2)4=a2(x22xy+y2)4=a2(xy)222 =(axay+2) (axay2) 30 (1)解:原式=(x3y) 2+4z(x3y)+4z2=(x3y+2z)2; (2)解:原式=(a 2+5a)2+10(a2+5a)+24120=(a2+5a)2+10(a2+5a)96 =(a 2+5a+16) (a2+5a6)=(a1) (a+6) (a2+5a+16)