F2 Maths (15)

係咪因式分解呀？
第１題用ａ＾２－ｂ＾２＝（ａ＋ｂ）（ａ－ｂ）　＜－呢條公式掛～
4a^2(y-2)^2-a^2
=a^2[4(y-2)^2-1] ＜－抽ａ２次出黎
=a^2[2(y-2)+1][2(y-2)-1]
=a^2(2y-4+1)(2y-4-1)
=a^2(2y-3)(2y+5) ＜－應該咁就完啦～～
第２題都係用第１條公式～
9a^2-4b^2-9a-6b
=(3a+2b)(3a-2b)-3(3a+2b)　＜－係-9a-6b到抽-3出黎～9a^2-4b^2咪用第１題ｇｅ公式囉～
=(3a+2b)(3a-2b-3) ＜－將(3a+2b)抽出黎～-3放入去(3a-2b)到～搞掂掛～
第３題都係用ｏ個條公式～
25(3x-2y)^2-9(2x+y)^2
=[5(3x-2y)+3(2x+y)][5(3x-2y)-3(2x+y)]
=(15x-10y+6x+3y)[15x-10y-(6x+3y)]
=(21x-7y)(15x-10y-6x-3y)
=(21x-7y)(9x-13y)
=7(3x-y)(9x-13y) <-如冇意外係咁做～希望我做岩啦～
ｐｓ只作參考～錯左唔關我事€

第一題：
[(2a)(y-2)]^2-a^2
＝[2a(y-2+1)][2a(y-2-1)]
=[2a(y-1)][2a(y-3)]

＊因為a２次－ｂ２次＝(a+b)(a-b)

第２題
9a^2-4b^2-9a-6b
=(3a)^2-(2b)^2-3(3a+2b)
=(3a+2b)(3a-2b)-3(2a+2b)
=(3a+2b)[3a-2b-3]


第３題
〔我唔抄題目〕
=[5(3X-2y)]^2 - [3(2x+y)]^2
=(15x-10y)^2 - (6x+3y)^2
=(15x-10y+6x+3y)[15x-10y-(6x+3y)]
=(21x-7y)(9x-13y)

希望冇計錯

1. 4a² (y-2)² - a²
= a² [ 2² (y-2)² - 1]　　　--- common factor : a²
= a² [ (2y-4 +1) (2y-4 -1)]　--- by x² - y² = (x+y) (x-y)
= a² (2y-3) (2y-5)

2. 9a² - 4b² -9a - 6b
= (3a)² - (2b)² - 3(3a) - 3(2b)
= (3a+2b) (3a-2b) - 3 (3a+2b)
= (3a+2b) (3a-2b-3)

3. 25 (3x-2y)² - 9 (2x+y)²
= 5² (3x-2y)² - 3² (2x+y)²
= [5 (3x-2y)+3 (2x+y)] [5 (3x-2y)-3(2x+y)]
= (15x-10y + 6x+3y) (15x-10y - 6x-3y)
= (21x-7y) (9x-13y)
= 7 (3x-y) (9x-13y)

我諗答案應該係咁

1. 4a² (y-2)² - a²
= a² [ 2² (y-2)² - 1]　　　--- common factor : a²
= a² [ (2y-4 +1) (2y-4 -1)]　--- by x² - y² = (x+y) (x-y)
= a² (2y-3) (2y-5)

2. 9a² - 4b² -9a - 6b
= (3a)² - (2b)² - 3(3a) - 3(2b)
= (3a+2b) (3a-2b) - 3 (3a+2b)
= (3a+2b) (3a-2b-3)

3. 25 (3x-2y)² - 9 (2x+y)²
= 5² (3x-2y)² - 3² (2x+y)²
= [5 (3x-2y)+3 (2x+y)] [5 (3x-2y)-3(2x+y)]
= (15x-10y + 6x+3y) (15x-10y - 6x-3y)
= (21x-7y) (9x-13y)
= 7 (3x-y) (9x-13y)