Question

初一因式分解

1. 4a²+4a-4a²b+b+1 2. a(a²-a-1)+1 3. ab(m²+n²)+mn(a²+b²)

Answer 1

1.原式=4a²+4a+1+b（1-4a²)
=（2a+1)²+b(1-2a)(1+2a)
=(2a+1)(2a+1+b-2ab)
2原式=a³-a²-a+1
=a²(a-1)-(a-1)
=(a²-1)(a-1)
=(a+1)(a-1)²
3.原式=abm²+abn²+mna²+mnb²
=ma(bm+na)+nb(na+mb)
=(bm+na)(ma+nb)

Answer 2

1. 4a²+4a-4a²b+b+1
=（4a²-4a²b）+4a+（b+1）
=4 (1-b) a²+4a+(b+1)
=[2(1-b)a+(b+1)] (2a+1)
2. a(a²-a-1)+1
=a3-a²-a+1
=a²(a-1)-(a-1)
=(a-1) (a²-1)
3. ab(m²+n²)+mn(a²+b²)
=abm²+abn²+mna²+mnb²
=mna²+(abm²+abn²)+mnb²
=mna²+(bm²+bn²)a+mnb²
=(ma+nb) (na+mb)

Answer 3

解： 4a2+4a-4a2b+b+1
=（2a+1）2-b(2a+1)(2a-1)
=（2a+1）[（2a+1）-b(2a-1)]
=（2a+1）（2a-2ab+b+1)
a(a2-a-1)+1
=a3-a2-a+1
=a2(a-1)-(a-1)
=(a-1)(a2-1)
=(a-1)2(a+1)
ab(m2+n2)+mn(a2+b2)
= abm2+abn2+mna2+mnb2
=am(bm+an)+bn(an+bm)
=(an+bm)(am+bn)