MATHS(f.2) only one question

49 − (2x −1)2
= [7 + (2x −1)][7 − (2x −1)]
= (2x + 6)(8 − 2x )

==> (2x + 6) = 2( x + 3)

==> (8 − 2x ) = 2 (4 − x )

= 2( x + 3) X 2 (4 − x )

= (2X2)( x + 3)(4 − x )

= 4( x + 3)(4 − x )

首先,49=7^2
然後,將條式看作
(7)^2 * (2x-1)^2

Then,用 formula a^2 * b^2 = (a+b)*(a-b) 做代入法
變做 [ 7+ (2x - 1) ] * [ 7- (2x - 1) ]
之後計......
(7+2x-1)(7-2x+1)
^依道要負負得正
=(2x+6)(8-x)
再抽 common factor (2)
=4[ (x+3) * (4-x) ]
=4 (x+3) (4-x)

明未?

2007-10-29 15:32:29 補充：
負負得正係[7-(2x-1)]果道

[7-(2x-1)][7+(2x-1)]
=(8-2x)(6+2x)
=2(4-x)2(3+x)
=4(4-x)(3+x)

49 − (2x −1)^2
=7^2-(2x-1)^2
=7^2-(2x)^2-2(2x 1)+1^2
=7^2-(2x)^2-4x+1
=(7+2x)(7-2x)-4x+1