1、(1)2x2x (2)16x21 (3)6xy29x2yy3 (4)4+12(xy)+9(xy)25因式分解:(1)2am28a (2)4x3+4x2y+xy26将下列各式分解因式:(1)3x12x3 (2)(x2+y2)24x2y27因式分解:(1)x2y2xy2+y3 (2)(x+2y)2y28对下列代数式分解因式:(1)n2(m2)n(2m) (2)(x1)(x3)+19分解因式:a24a+4b2 10分解因式:a2b22a+111把下列各式分解因式:(1)x47x2+1 (2)x4+x2+2ax+1a2(3)(1+y)22x2(1y2)+x4(1y)2 (4)x4+2x3+3x2+2x
2、+112把下列各式分解因式:(1)4x331x+15; (2)2a2b2+2a2c2+2b2c2a4b4c4; (3)x5+x+1;(4)x3+5x2+3x9; (5)2a4a36a2a+2解答:解:(1)3p26pq=3p(p2q),(2)2x2+8x+8,=2(x2+4x+4),=2(x+2)2 (1)x3yxy (2)3a36a2b+3ab2 分析:(1)首先提取公因式xy,再利用平方差公式进行二次分解即可;(2)首先提取公因式3a,再利用完全平方公式进行二次分解即可(1)原式=xy(x21)=xy(x+1)(x1);(2)原式=3a(a22ab+b2)=3a(ab)2(1)a2(xy)
3、+16(yx); (2)(x2+y2)24x2y2(1)a2(xy)+16(yx),=(xy)(a216),=(xy)(a+4)(a4);(2)(x2+y2)24x2y2,=(x2+2xy+y2)(x22xy+y2),=(x+y)2(xy)2(1)2x2x; (2)16x21; (3)6xy29x2yy3; (4)4+12(xy)+9(xy)2(1)2x2x=x(2x1);(2)16x21=(4x+1)(4x1);(3)6xy29x2yy3,=y(9x26xy+y2),=y(3xy)2;(4)4+12(xy)+9(xy)2,=2+3(xy)2,=(3x3y+2)2(1)2am28a; (2)4
4、x3+4x2y+xy2(1)2am28a=2a(m24)=2a(m+2)(m2);(2)4x3+4x2y+xy2,=x(4x2+4xy+y2),=x(2x+y)2(1)3x12x3 (2)(x2+y2)24x2y2(1)3x12x3=3x(14x2)=3x(1+2x)(12x);(2)(x2+y2)24x2y2=(x2+y2+2xy)(x2+y22xy)=(x+y)2(xy)2(1)x2y2xy2+y3; (2)(x+2y)2y2(1)n2(m2)n(2m); (2)(x1)(x3)+1a24a+4b2 10分解因式:(1)x47x2+1; (2)x4+x2+2ax+1a2(3)(1+y)22
5、x2(1y2)+x4(1y)2 (4)x4+2x3+3x2+2x+1(1)x47x2+1=x4+2x2+19x2=(x2+1)2(3x)2=(x2+3x+1)(x23x+1);(2)x4+x2+2ax+1a=x4+2x2+1x2+2axa2=(x2+1)(xa)2=(x2+1+xa)(x2+1x+a);(3)(1+y)22x2(1y2)+x4(1y)2=(1+y)22x2(1y)(1+y)+x4(1y)2=(1+y)22x2(1y)(1+y)+x2(1y)2=(1+y)x2(1y)2=(1+yx2+x2y)2(4)x4+2x3+3x2+2x+1=x4+x3+x2+x3+x2+x+x2+x+1=
6、x2(x2+x+1)+x(x2+x+1)+x2+x+1=(x2+x+1)2 (2)2a2b2+2a2c2+2b2c2a4b4c4;(3)x5+x+1; (4)x3+5x2+3x9;(5)2a4a36a2a+2(1)4x331x+15=4x3x30x+15=x(2x+1)(2x1)15(2x1)=(2x1)(2x2+115)=(2x1)(2x5)(x+3);(2)2a2b2+2a2c2+2b2c2a4b4c4=4a2b2(a4+b4+c4+2a2b22a2c22b2c2)=(2ab)2(a2+b2c2)2=(2ab+a2+b2c2)(2aba2b2+c2)=(a+b+c)(a+bc)(c+ab)(ca+b);(3)x5+x+1=x5x2+x2+x+1=x2(x31)+(x2+x+1)=x2(x1)(x2+x+1)+(x2+x+1)=(x2+x+1)(x3x2+1);(4)x3+5x2+3x9=(x3x2)+(6x26x)+(9x9)=x2(x1)+6x(x1)+9(x1)=(x1)(x+3)2;(5)2a4a36a2a+2=a3(2a1)(2a1)(3a+2)=(2a1)(a33a2)=(2a1)(a3+a2a2a2a2)=(2a1)a2(a+1)a(a+1)2(a+1)=(2a1)(a+1)(a2a2)=(a+1)2(a2)(2a1)
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