f.2 expand 1題 急.!!

(-a+b)(-a-b)

-(-a+b)(a+b)
(a-b)(a+b)
a^2-b^2
____________
((-2p+1)+q))((-2p+1)-q))

(-2p+1)^2-q^2

1-4p+4p^2-q^2
____________________

2010-10-24 17:29:22 補充：
.

1.
(-a+b)(-a-b)
=[-(a-b)][-(a+b)]
=(a-b)(a+b)
=a^2-b^2

2.
(-2p+1+q)(-2p+1-q)
=[-(2p-1)+q][-(2p-1)-q]............. (3)
and now let -(2p-1) = a and q = b and sub into (3)

[-(2p-1)+q][-(2p-1)-q]
=[-(2p)]^2 - q^2
= 4p^2 - q^2



2010-10-24 18:44:02 補充：
[-(2p-1)+q][-(2p-1)-q]
=[-(2p)]^2 - q^2
= 4p^2 - q^2 更正.....


[-(2p-1)+q][-(2p-1)-q]
=[-(2p-1)]^2 - q^2
= (2p-1)^2 - q^2
= 4p^2 - 4p - 1 - q^2

2010-10-24 18:46:07 補充：
sorry... 最後更正....

1.
(-a+b)(-a-b)
=[-(a-b)][-(a+b)]
=(a-b)(a+b)
=a^2-b^2

2.
(-2p+1+q)(-2p+1-q)
=[-(2p-1)+q][-(2p-1)-q]............. (3)
and now let -(2p-1) = a and q = b and sub into (3)

[-(2p-1)+q][-(2p-1)-q]
=[-(2p-1)]^2 - q^2
=(2p-1)^2 - q^2
= 4p^2 - 4p + 1 - q^2

** x^2 = (x)(x) , 3^2 = 3 x 3 = 9 **

1.expand (-a+b)(-a-b)

Do you remember that : x^2 - y^2 = ( x + y ) ( x - y ) ?

So , let -a = x , b = y

( -a + b ) ( -a - b )

= ( x + y ) ( x - y )

= x^2 - y^2

= ( -a ) ^2 - b^2

= a^2 - b^2

2.by using the result of (1) ,expand (-2p+1+q)(-2p+1-q)

Let -2p + 1 = a , q = b

( -2p + 1 + q ) ( -2p + 1 - q )

= ( a + b ) ( a - b )

= a^2 - b^2

= ( -2p + 1 ) ^2 - q^2

= -2p ( -2p + 1 ) + 1 ( -2p + 1 ) - q^2

= -2p ( -2p ) - 2p - 2p + 1 - q^2

= 2p ( 2p ) - 4p + 1 - q^2

= 4p^2 - 4p + 1 - q^2

= 4p^2 - q^2 - 4p + 1

1.(-a+b)(-a-b)
=(-a)(-a-b)+b(-a-b)
=a^(2)+ab-ab-b^(2)
=a^(2)-b^(2) <<<<這裡也可以直接用恆等式

2.Let (-a) be (-2p+1) and (b) be (q).
(-2p+1+q)(-2p+1-q)
=(-a+b)(-a-b)
=a^(2)-b^(2)
=(-2p+1)^(2)-(q)^(2)
=[4p^(2)-4p+1]-q^(2)
=4p^(2)-4p+1-q^(2)

***** ^(2)即是2次方