math 問題(20)

(a-b)^2+(a+b) <> (a-b)(a+b)+(a+b), 即係唔等於

(a-b)^2+(a+b) = (a-b)(a-b) + (a+b)
= a(a-b) - b(a-b) + a + b
= a^2 - ab - ab + b^2 + a + b
= a^2 - ab + a + b^2 - ab + b
= a^2 - 2ab + b^2 + a + b
= a^2 - 2ab + b^2 + a + b ...... 係冇得簡化嘅.



正常來說,(a-b)^2 = ?? ..... 數學是沒有正常不正常的, 只有一個答案; 當然或會因為應用條件不同而最終會選擇或排除某些答案; 但條件相同, 結果永遠是一樣的.

a^2+b^2 就等於 a^2+b^2, 冇得再簡化.



(a-b)^3 - (c+d)^3
= (a-b)^3 + [- (c+d)^3]
= (a-b)^3 + [(-1)^3 * (c+d)^3]
= (a-b)^3 + [(-1) * (c+d)]^3
= (a-b)^3 + [- (c+d)]^3
= (a-b)^3 + (-c-d)^3
= [(a-b) + (-c-d)] [(a-b)^2 - (a-b)(-c-d) + (-c-d)^2] ........ 因為 a^3 + b^3 = (a+b)(a^2-ab+b^2)
= (a-b-c-d) [a^2 - 2ab + b^2 - (-ac-ad+bc+bd) + c^2 + 2cd + d^2]
= (a-b-c-d) (a^2 - 2ab + b^2 + ac + ad - bc - bd + c^2 + 2cd + d^2)
= (a-b-c-d) (a^2 + b^2 + c^2 + d^2 - 2ab + ac + ad - bc - bd + 2cd)
= a^3 + ab^2 + ac^2 + ad^2 - 2a^2b + a^2c + a^2d - abc - abd + 2acd
- a^2b - b^3 - bc^2 - bd^2 + 2ab^2 - abc - abd + b^2c + b^2d - 2bcd
- a^2c - b^2c - c^3 - cd^2 + 2abc - ac^2 - acd + bc^2 + bcd - 2c^2d
- a^2d - b^2d - c^2d - d^3 + 2abd - acd - ad^2 + bcd + bd^2 - 2cd^2
= a^3 - b^3 - c^3 - d^3 - 3a^2b + 3ab^2 - 3c^2d - 3cd^2

2007-11-06 20:32:28 補充：
a^2+b^2 就等於 a^2+b^2, 冇得再簡化a^2 - b^2 = (a - b) (a + b)(a - b)^2 = a^2 - 2ab + b^2(a + b)^2 = a^2 + 2ab + b^2請小心看清楚括號位置及+/- 號.

(a-b)^2+(a+b) <<<<唔係
是否 = (a-b)(a+b)+(a+b) a^2-b^2先係(a+b)(a-b)
and then =?

正常來說,(a-b)^2 = ?? (a-b)^2即係a^2-2ab+b^2
a^2+b^2 = ??