Question

分解因式：a²-b²-2a+1=? 还有一题:(x²-5)²+8(5-x²)+16 具体过程，谢谢

Answer 1

1.a²-b²-2a+1=(a-1)^2-b^2=(a-1+b)*(a-1-b)

2.(x²-5)²+8(5-x²)+16=(5-x²+4)^2=(x²-9)^2=(x-3)^2*(x+3)^2

Answer 2

a²-b²-2a+1
=（a-1）²-b²
=（a-1-b）（a-1+b）

（x²-5)²+8(5-x²)+16
=（x²-5)²-8(x²-5)+16
=[（x²-5）-4]²
=（x²-9）²
=[（x-3)(x+3）]²

Answer 3

a²-b²-2a+1=a²-2a+1-b²=(a-1)(a-1)-b²=(a-1+b)(a-1-b)

设5-x²=a
(x²-5)²+8(5-x²)+16=a²+8a+16=(a+4)(a+4)=(5-x²+4)(5-x²+4)=(9-x²)(9-x²)
=(3-x)(3-x)(3+x)(3+x)

Answer 4

a�0�5-b�0�5-2a+1=（a�0�5-2a+1）-b�0�5=(a-1)�0�5-b�0�5=(a-1-b)(a-1+b)
令a=(x�0�5-5)则(x�0�5-5)�0�5+8(5-x�0�5)+16=a�0�5-8a+16=(a-4)�0�5=(x�0�5-9)�0�5=
(x-3)�0�5(x+3)�0�5