F2. Identities and Factor....

2012-09-20 3:13 am

Perfect squares (Subtraction)

Previous method
eg.) (a-b)^2
= (a - b)(a - b)
= a^2 - ab - ba + b^2

Faster method
( a - b )^2
= ( )^2( )( )+( )^2
=_______________________

1. Previous method
( x - y)^2
= (x - y)(x - y)
=_____________
=_____________

Faster method
(x - y)^2
= ( )^2-2( )( )+( )^2
= ____________
=_____________

2. Previous method
(5a - 7)^2
=____________
=____________
=____________
Faster method
(5a - 7)^2
=( )^2( )( )+( )^2
=__________________

3. Previous method
(pq - 5)^2

Faster method
(pq - 5)^2
= ( )^2-( )( )+( )^2
=_________________

另外圖:
http://upload.lsforum.net/users/public/v40833435346d216.jpg

THANK~

回答 (1)

50418129

2012-09-20 10:26 pm

✔ 最佳答案

eg.) Previous method
(a - b)^2
= (a - b)(a - b)
= a^2 - ab - ba + b^2

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

1. Previous method
(x - y)^2
= (x - y)(x - y)
= x^2 - xy - yx + y^2
= x^2 - 2xy + y^2

Faster method
(x - y)^2
= (x)^2 - 2(x)(y) + (y)^2
= x^2 - 2xy + y^2

2. Previous method
(5a - 7)^2
= (5a - 7)(5a - 7)
= (5a)^2 - (5a)(7) - (7)(5a) + 7^2
= 25a^2 - 70a + 49

Faster method
(5a - 7)^2
=(5a)^2 - 2(5a)(7) + (7)^2
= 25a^2 - 70a + 49

3. Previous method
(pq - 5)^2
= (pq - 5)(pq - 5)
= (pq)^2 - (pq)(5) - (5)(pq) + 5^2
= p^2q^2 - 10pq + 25

Faster method
(pq - 5)^2
= (pq)^2- 2(pq)(5) + (5)^2
= p^2q^2 - 10pq + 25

參考: knowledge

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