Help!!! 有 maths 唔識 ... 恆等式&因式分解

1.)

Let a = x + y

64 - (x + y)2

= 82 - a2

= [8 + a][8 - a]

= [8 + (x + y)][8 - (x + y)]

= (8 + x + y)(8 - x - y)

=====
2.)
(x2 - 6xy + 9y2) - x2

= [x2 - 2x(3y) + (3y)2] - x2

= (x - 3y)2 - x2

= [(x - 3y) - x][(x - 3y) + x]

= (x - 3y - x)(x - 3y + x)

= -3y(2x - 3y)

or 3y(3y - 2x)

=====
3.)
Let a = 3x - y


(3x - y)2 + 4(3x - y) + 4

= a2 - 4a + 4

= a2 - 2(a)(2) + (2)2

= [a - 2]2

= [(3x - y) - 2]2

= (3x - y - 2)2

=====
4.)

(x2 + 2x + 1) + xy + y

= (x2 + 2x + 1) + (xy + y)

= (x + 1)2 + (x + 1)y

= (x + 1)[(x + 1) + y]

= (x + 1)(x + y + 1)
=

1. 64 - (x+y) 二次方
=64-(x二次方+y二次方)
=64-x二次方-y二次方

2. (x二次方 - 6xy + 9y二次方 ) - x二次方
=( x二次方 - 6xy + 9y二次方 ) - x二次方
= x二次方 - 6xy +9 y二次方 - x二次方
= x二次方 - x二次方 - 6xy + 9y二次方
= 9y二次方 -6xy

p.s. 9y二次方 不等於(9y)二次方
9y二次方=9y二次方
(9y)二次方=81y二次方

3. (3x-y)二次方+4(3x-y)+4
=(9x二次方-y二次方)+(12x-4y)+4
=9x二次方-y二次方+12x-4y+4

4. (x二次方+2x+1)+xy+y
x二次方+2x+1+xy+y