分类：教育/科学-学习帮助
问题：因式分解3~~~a^5
a^5-a^2*b^3-a^3*b^2+b^5
x^4+4
x^2-5xy+6y^2-3x+6y
20x^2-39x+18
(x+1)(x+2)(x+3)(x+4)-120
7x^2-3y+xy-21x
(x^2+3x-3)(x^2+3x+4)-8
(x+1)(x+2)(x+3)(x+4)-24
a^4+1/4
x^4+x^2*y^2+y^4
-12a^2m+1+b^m+2+2a^m+1*b^2m+4
我要过程
拜托啦！实在不会写！
回答：　a^5-a^2*b^3-a^3*b^2+b^5
= (a^5-a^2*b^3)-(a^3*b^2-b^5)
= a^2*(a^3-b^3)-b^2*(a^3-b^3)
= (a^2-b^2)(a^3-b^3)
= (a-b)^2*(a+b)*(a^2+ab+b^2)

　x^2-5xy+6y^2-3x+6y
= (x-2y)(x-3y)-3(x-2y)
= (x-2y)(x-3y-3)

20x^2-39x+18 = (4x-3)(5x-6)

　(x+1)(x+2)(x+3)(x+4)-120
= [(x+1)(x+4)][(x+2)(x+3)]-120
= (x^2+5x+4)(x^2+5x+4+2)-120
= (x^2+5x+4)^2 + 2(x^2+5x+4) - 120
= (x^2+5x+4-10)(x^2+5x+4+12)
= (x-1)(x+6)(x^2+5x+16)

　7x^2-3y+xy-21x
= (7x^2+xy)-(3y+21x)
= (7x+y)(x-3)

　(x^2+3x-3)(x^2+3x+4)-8
= (x^2+3x)^2 + (x^2+3x) - 20
= (x^2+3x-4)(x^2+3x+5)
= (x-1)(x+4)(x^2+3x+5)

　(x+1)(x+2)(x+3)(x+4)-24
= [(x+1)(x+4)][(x+2)(x+3)]-24
= (x^2+5x+4)(x^2+5x+4+2)-24
= (x^2+5x+4)^2 + 2(x^2+5x+4) - 24
= (x^2+5x+4-4)(x^2+5x+4+6)
= x(x+5)(x^2+5x+10)

　x^4+x^2*y^2+y^4
= (x^4+2x^2*y^2+y^4)-(xy)^2
= (x^2+y^2)^2 - (xy)^2
= (x^2+xy+y^2)(x^2-xy+y^2)

-12a^(2m+1)*b^(m+2)+2a^(m+1)*b^(2m+4)
= 2a^(m+1)*b^(m+2)*[-6a^m+b^(m+2)]