f.2 恆等式, (a-b)^2

(a-b)^2
=(a-b)(a-b)
=a^2-ab-ab+b^2
=a^2-2ab+b^2

& -(a-b)b-b(a-b)
=-(a-b)b-(a-b)b
=-(a-b)(b+b)
=-2b(a-b)

2010-10-11 06:53:00 補充：
(a-b)^2
=(a-b)(a-b)
=a(a-b) -b(a-b)
=a^2-ab-(ab-b^2)
=a^2-ab-ab+b^2
=a^2-2ab+b^2

2010-10-11 21:16:13 補充：
[sorry,打錯,係(a-b)^2=a^2-2ab+b^2先岩
我係吾明a^2-(a-b)b-b(a-b)-b^2
=a^2-b^2-2b(a-b)
=a^2-2ab+b^2呢題姐,
(a-b)^2=a^2-2ab+b^2同(a-b)^2(a-b)^2 = ( a - b ) ( a - b )呢D明嫁la
[second list wrong:
(a-b)^2(a-b)^2=(a-b)^4]
-(a-b)b-(a-b)b點解變左-(a-b)(b+b) ←仲係吾明= =<--001 wrong,it's his problem

(a-b)(a-b)=a(a-b) -b(a-b)<--

2010-10-12 06:43:24 補充：
remind u:
not I said t.t

(a-b)^2
(a)^2-2(a)(b)+(b)^2
a^2-2ab+b^2
_____________
For example:
(xy-ab)^2
(xy)^2-2(xy)(ab)+(ab)^2
x^2y^2-2xyab+a^2b^2
___________________
Remind you:
(a-b)^2=a^2-2ab+b^2=b^2-2ab+a^2
(b-a)^2=b^2-2ab+a^2

(a-b)^2=(b-a)^2

(a-b)^2

= ( a - b ) ( a - b )

= a ( a - b ) - b ( a - b )

= a^2 - ab - ( ab - b^2 )

= a^2 - ab - ab + b^2

= a^2 - 2ab + b^2