F.2 數學問題(聽日考救急)

11a) (x + 3)²
= x² + 2x(3) + (3)²
= x² + 6x + 9
b) Hence ,
(x+2y + 3)²
= (x+2y)² + 6(x+2y) + 9 ....... (By the result of part a)
= x² + 2x(2y) + (2y)² + 6x + 12y + 9
= x² + 4xy + 4y² + 6x + 12y + 9

2012-10-30 23:26:51 補充：
把 x+2y 當成 part a) 個 x 咁計~

11.
(a) Expand (x+3)^2
(x+3)^2
= x^2 +6x +9

(b) Hence, expand (x+2y+3)^2
2 methods here
Method 1
(x+2y+3)^2
= [ (x+2y) +3]^2
= (x+2y)^2 +6(x+2y) +9
= (x^2 +4xy + 4y^2) + 6x +12y + 9
= x^2 +4xy + 4y^2 + 6x +12y + 9

Method 2
(x+2y+3)^2
= [ (x+3) +2y]^2
= (x+3)^2 +2(x+3)(2y) +4y^2
= x^2 +6x +9 + 2(2xy+6y) +4y^2
= x^2 +4xy + 4y^2 + 6x +12y + 9

both accept

prefer Method 1