Question

求解几道初一的因式分解题，谢谢

刚开始学，有些题目很灵活，所以求解

(x+y)(x^2-y^2+y-x) 做到一半，不知道后面一个括号该怎么提

（3a+2b)^2-4(a-b)^2

16(2m+n)^2-9(m-2n)^2

(x+y)^2-2(x^2-y^2)+(x-y)^2

谢谢

Answer 1

1.(x+y)(x²-y²+y-x)
=(x+y)[(x²-y²)-(x-y)]
=(x+y)[(x+y))(x-y)-(x-y)]
=(x+y)(x-y)(x+y-1)
2.(3a+2b)²-4(a-b)^2
=[(3a+2b)+2(a-b)][(3a+2b)-2(a+b)]
=(3a+2b+2a-2b)(3a+2b-2a+2b)
=5a(a+4b)
3.16(2m+n)^2-9(m-2n)^2
=[4(2m+n)]²-[3(m-2n)]²
=[(4(2m+n)+3(m-2n)][4(2m+n)-3(m-2n)]
=[8m+4n+3m+6n][8m+4n-3m+6n]
=(11m-2n)(5m+10n)
=5(11m-2n)(m+2n)
5.(x+y)^2-2(x^2-y^2)+(x-y)^2
=(x+y)²-2(x+y)(x-y)+(x-y)²
=[(x+y)-(x-y)]²
=4y²

以上题目中均用到了数学的整体思维，1~4题还用了平方差公式和乘法分配律，第5题用到了完全平方公式

Answer 2

x+y)(x^2-y^2+y-x)
=(x+y)[(x²-y²)-(x-y)]
=(x+y)(x-y)(x+y-1)

(3a+2b)²-4(a-b)^2
=(3a+2b+2a-2b)(3a+2b-2a+2b)
=5a(a+4b)

16(2m+n)^2-9(m-2n)^2
=[(4(2m+n)+3(m-2n)][4(2m+n)-3(m-2n)]
=(11m-2n)(5m+10n)
=5(11m-2n)(m+2n)

(x+y)^2-2(x^2-y^2)+(x-y)^2
=[(x+y)-(x-y)]²
=4y²