Question

初二数学，两道因式分解，求学霸。（过程尽量详细）

Answer 1

1、原式=a^2(b-1)+b^2(b+1)
=(b+1)(a^2+b^2)

2、原式=(3m^2-n^2+m^2-3n^2)(3m^2-n^2-m^2-3n^2)
=(4m^2-4n^2)(2m^2-4n^2)
=4*2(m+n)(m-n)(m+√2n)(m-√2n)
=8(m+n)(m-n)(m+√2n)(m-√2n)

Answer 2

a²(b-1)+b²-b³
=a²(b-1)-b²(b-1)
=(b-1)(a²-b²)
=(b-1)(a+b)(a-b)

(3m²-n²)²-(m²-3n²)²
=[(3m²-n²)+(m²-3n²)][(3m²-n²)-(m²-3n²)]
=(3m²-n²+m²-3n²)(3m²-n²-m²+3n²)
=(4m²-4n²)(2m²+2n²)
=8(m²+n²)(m+n)(m-n)

Answer 3

（a+b)(a-b)(b-1)
=(3m2-n2+m2-3n2)(3m2-n2-m2+3n2)
=4(m+n)(m-n)2(m2+n2)
=8(m+n)(m-n)(m2+n2)

Answer 4

知道

Answer 5

（a^2+b^2)*（b-1）