一些初2因式分解的题
问题描述：
一些初2因式分解的题
计算 3a³b²÷a²+b×(a²b-3ab-5a²b)
化简 1.y(x+y)+(x+y)(x-y)-x²
2.2a²-a(a-3)-3[a²-5a(-a-2)
3.3(m+1)²-5(m+1)(m-1)+2(m-1)²-2mn
4.[(x-y)²+(x+y)(x-y)]÷2x
答
3a³b²÷a²+b×(a²b-3ab-5a²b)
=3ab²+a²b²-3ab²-5a²b²
=-4a²b²
1,y(x+y)+(x+y)(x-y)-x²
=(y+x-y)(x+y)-x²
=x²+xy-x²
=xy
2,2a²-a(a-3)-3[a²-5a(-a-2)]
=2a²-a²+3a-3a²-15a²-30a
=-17a²-27a
3,3(m+1)²-5(m+1)(m-1)+2(m-1)²-2mn
=-[(m+1)+(m-1)][3(m+1)+2(m-1)]-2mn
=-2m(5m+1)-2mn
=-10m²-2m-2mn
4,[(x-y)²+(x+y)(x-y)]÷2x
=(x-y)[(x-y)+(x+y)]÷2x
=2x(x-y)÷2x
=x-y