f.2 maths

1. 3a²b² ×7a² b ³
= 3 x 7 x a^(2+2) x b (2+3)
= 21a^4b^5
2. ab²(a²-b³)
= ab² x a² - ab² x b³
= a^(1+2)b^2 - ab^(2+3)
=a³b² - ab^5

(3y+4)(2y-3)
= 3y x 2y +4 x 2y +(-3 x 3y) + (4 x -3)
= 6y² - 9y + 8y -12
= 6y² - y -12

3.
ab³ -6b²
= b² (ab-6) (大家都有b²)

15xy²-9x²y
= 3xy (5y-3x)

a²(a-b)-b³ (b-a)
= a²(a-b)+b³ (a-b) (將符號從a和b改變，在b³一項改變連帶符號)
= (a-b)(a²+b³)

ab-2b+a-2
= ab+a -2-2b (將含公因數的項放在一起)
= a(b+1) -2 (1+b)
= (a-2)(b+1)

4. 1/2(a-2)(a+2)
= 1/2 (a²-2²) (identity (a-b)(a+b)= a²-b²)
= 1/2 (a²- 4)

第五題表示得不太清楚...以下是我的推測...
5. 2(1-2X)-A = Bx+2
L.H.S. = 2-4x-A
R.H.S. = Bx+2
Therefore A = 0
B = -4
(x-2)²+A=x²+BX+2
L.H.S. = x²-4x+4+A
R.H.S. = x²+BX+2

Therefore A=-2
B = -4

1) 3a²b² ×7a² b ³ => 3 x 7 x a(2+2) x b(2+3) => 21a4b5
2) ab²(a²-b³) = aa2b2 – ab2b3 = a(1+2) b2 – a b(2+3) = a3b2 – ab5
(3y+4)(2y-3) = (3y x 2y) – (3y x 3) + (4 x 2y) – (4 x 3) = 6y2 – 9y + 8y - 12 = 6y2 – y - 12
3)ab³ -6b² = abb2 -6b² = b2 (ab-6)
15xy²-9x²y = (3*5)xyy-(3*3)xxy = (5y – 3x)3xy
a²(a-b)-b³ (b-a) = a²(a-b)- (-b³ (-b+a)) = a²(a-b) + b³ (a-b) = (a²+ b³ )(a-b)
ab-2b+a-2 = ab+a – 2(b+1) = a(b+1) – 2(b+1) = (a-2)(b+1)
4) 1/2(a-2)(a+2) = ½ (a2- 2a + 2a - 22) = ½ (a2 - 22) = ½ a2 - 2
5) 問題有問題