Question

七年级因式分解

1.a(x²-2x)²-2ax(2-x)+a
2.简便计算：201²-52²+253×851
3.三分之二b-九分之一b²
4.(ax+by)²-2acxz-2bcyz+c²z²
5.(a+2b)四次方-（a-2b）四次方
要过程

Answer 1

1.原式=a[(x^2 - 2x)^2 - 2x(2-x) + 1]
=a[(x^2 - 2x)^2 + 2(x^2 - 2x) + 1]
=a[(x^2 - 2x) + 1]^2
=a(x^2 - 2x + 1)^2
=a(x-1)^4

2.原式=(201+52)(201-52) + 253×851
=253×149 + 253×851
=253×(149 + 851)
=253×1000
=253000

3.原式=(b/9)(6-b)

4.原式=(ax+by)^2 - 2cz(ax+by) + c^2z^2 =(ax + by - cz)^2

5.原式=[(a+2b)^2 + (a-2b)^2][(a+2b)^2 - (a-2b)^2]
=[(a+2b)^2 + (a-2b)^2][(a+2b)+(a-2b)][(a+2b)-(a-2b)]
=(a^2 + 4ab + 4b^2 + a^2 - 4ab + 4b^2)×(2a)×(4b)
=2(a^2 + 4b^2)×8ab
=16ab(a^2 + 4b^2)

Answer 2

1 a(x²-2x)²-2ax(2-x)+a
=a[(x²-2x)²-2x(2-x)+1]
=a[(x²-2x)²+2x(x-2)+1]
=a(x²-2x+1)²
=a(x-1)

2 201²-52²+253×851
=(201+52)(201-52)+253×851
=253×149+253×851
=253×（149+851）
=253000

3 (2b)/3-b²/9
=b/9(6-b)

4 (ax+by)²-2acxz-2bcyz+c²z²
=(ax+by)²-2cz(ax+by)+c²z²
=(ax+by-cz)²

5 (a+2b)⁴－（a-2b)⁴
=[(a+2b)²-(a-2b)²][(a+2b)²+（a-2b)²]
=16ab(a²+4b²)

Answer 3

2.201²-52²+253×851
=(201+52)(201-52)+253×851
=253×149+253×851
=253×(149+851)
=253×1000
=253000
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