a-maths!!!!!!

用戶7639

2007-09-25 1:46 am

(a)証明(a+b)⁴-4ab(a+b)²+2a²b²=a⁴+b⁴。
(b)若a為正數,b為負數,且a、b滿足a+b=1及a⁴+b⁴=97,利用(a)結果,証明ab=-6。
(c)由此求a和b的值。

回答 (1)

Audrey Hepburn

2007-09-25 1:56 am

✔ 最佳答案

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

參考: Myself~~~

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