Question

问几道关于因式分解的题目 （1）已知a^2b^2+a^2+b^2+10ab+16=0 (2)5a^2b(x-y)^2-1？

Answer 1

（1）已知a^2b^2+a^2+b^2+10ab+16=0
a^2b^2+a^2+b^2+10ab+16=0 则
a^2b^2+8ab+16+a^2+b^2+2ab=0
(ab+4)^2+(a+b)^2=0
因此 |ab+4|=0
|a+b|=0 a=+-2 b=-+2
(2)5a^2b(x-y)^2-10ab^2(y-x^)3
=5ab(x-y)^2(a+2b（x-y))=5ab(x-y)^2(a+2bx-2by)
( 3)3分之1a^2-3b^2
=1/3（a^2-9b^2）=1/3（a-3b)(a+3b)
(4)-4x^2-9y^2-30xy
(5)(x+y)^3-(x-y)^3
=[(x+y)-(x-y)][(x+y)^2+(x+y)(x-y)+(x-y)^2]
=2y[(x+y)(x+y+x-y)+(x-y)^2]
=2y[2x(x+y)+(x-y)^2]
=2y[2x^2+2xy+x^2-2xy+y^2]
=2y(3x^2-y^2)
(6)9a^2+x^2n+6a+2x^n+6ax^n+1
解 =9a^2+6ax^n+ x^2n+6a+2x^n+1
= (3a+x^n)^2+2(3a+x^n)+1
=(x^n+2a)^2,8,第五题利用立方差公式(x+y)^3-(x-y)^3 = 2y(3x^2+y^2)

第三题1/3a^2-3b^2=3[(1/3a)^2-b^2]=3（1/3a+b）（1/3a-b）,0,问几道关于因式分解的题目
（1）已知a^2b^2+a^2+b^2+10ab+16=0 (2)5a^2b(x-y)^2-10ab^2(y-x^)3 ( 3)3分之1a^2-3b^2 (4)-4x^2-9y^2-30xy (5)(x+y)^3-(x-y)^3 (6)9a^2+x^2n+6a+2x^n+6ax^n+1