標題:
a-maths!!!!!!
發問:
最佳解答:
a. L.H.S. = (a + b)4 – 4ab(a + b)2 + 2a2b2 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4) – 4ab(a2 + 2ab + b2) + 2a2b2 (binomial theorem) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 – 4a3b – 8a2b2 – 4ab3 + 2a2b2 = a4 + b4 = R.H.S. b. a > 0, b < 0 a + b = 1 a4 + b4 = 97 i.e. (a + b)4 – 4ab(a + b)2 + 2a2b2 = 97 (from a) (1)4 – 4ab(1)2 + 2(ab)2 = 97 2(ab)2 – 4ab – 96 = 0 (ab)2 – 2ab – 48 = 0 (ab - 8)(ab + 6) = 0 ab = 8 (rejected, since a > 0, b < 0, ab < 0) or -6 So, ab = -6 For a + b = 1, ab = -6, a4 + b4 = 97 The only possibility which satisfy the condition is: a = 3, b = -2
其他解答:
a-maths!!!!!!
發問:
此文章來自奇摩知識+如有不便請留言告知
(a)証明(a+b)?-4ab(a+b)2+2a2b2=a?+b?。 (b)若a為正數,b為負數,且a、b滿足a+b=1及a?+b?=97,利用(a)結果,証明ab=-6。 (c)由此求a和b的值。最佳解答:
a. L.H.S. = (a + b)4 – 4ab(a + b)2 + 2a2b2 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4) – 4ab(a2 + 2ab + b2) + 2a2b2 (binomial theorem) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 – 4a3b – 8a2b2 – 4ab3 + 2a2b2 = a4 + b4 = R.H.S. b. a > 0, b < 0 a + b = 1 a4 + b4 = 97 i.e. (a + b)4 – 4ab(a + b)2 + 2a2b2 = 97 (from a) (1)4 – 4ab(1)2 + 2(ab)2 = 97 2(ab)2 – 4ab – 96 = 0 (ab)2 – 2ab – 48 = 0 (ab - 8)(ab + 6) = 0 ab = 8 (rejected, since a > 0, b < 0, ab < 0) or -6 So, ab = -6 For a + b = 1, ab = -6, a4 + b4 = 97 The only possibility which satisfy the condition is: a = 3, b = -2
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