求解~详细过程~因式分解~拜托

Answer 1

1原式=x^2(x^2-2x-35)=x^2(x-7)(x+5)
3、原式=（5x-10y)^2-(4y-2x)^2
=(5x-10y+4y-2x)(5x-10y-4y+2x)
=(3x-6y)(7x-14y)
=21(x-2y)^2
5、原式=x(x^4-1)=x(x^2+1)(x^2-1)=x(x^2+1)(x-1)(x+1)
7、原式=(a-b)x^2-(a-b)x-(a-b)=(a-b)(x^2-x-1)
9、原式=9(x^4-4y^2)=9(x^2+2y)(x^2=2y)

满意就采纳吧 ，谢谢！

Answer 2

x^4-2x^3-35x^2
=x^2(x^2-2x-35)
=x^2(x-7)(x+5)

25(x-2y)^2-4(2y-x^2)
=[5(x-2y)]^2-[2(2y-x)]^2
=[5(x-2y)+2(2y-x)][5(x-2y)-2(2y-x)]
=(3x-6y)(7x-14y)
=21(x-2y)^2

x^5-x
=x(x^4-1)
=x(x^2+1)(x^2-1)
=x(x^2+1)(x+1)(x-1)

ax^2-bx^2-bx+ax+b-a
=a(x^2+x-1)-b(x^2+x-1)
=(a-b)(x^2+x-1)

9x^4-36y^2
=(3x^2)^2-(6y)^2
=(3x^2+6y)(3x^2-6y)
=9(x^2+2y)(x^2-2y)

Answer 3

1,原式=x²（x²-2x-35）=x²（x-7)(x+5)
3,原式=(x-2y)²（25-4）=21(x-2y)²
5，原式=x(x四次方-1）=x(x²+1）（x²-1）=x(x²+1)(x+1)(x-1)
7,原式=（a-b)x²+（a-b)x-(a-b)=(a-b)(x²+x-1）
9，原式=9(x四次方-4y²=9（x²+2y）（x²-2y)
ps：打的手累死了，望采纳