al pure maths substitution.123

2012-03-07 7:13 am
http://i987.photobucket.com/albums/ae355/delaynopemore/2-11.jpg
更新1:

我想問 c 的 substitution 應該係咩? a, b 我都諗到

更新2:

岩哂啊! 但我都係唔知 c = 2x + 8√2 - 12√3 係點觀察出黎, 係要爆開一d野先, 定....? a = x - 2√2 + 5√3, b = -3x - 6√2 + 7√3 都好容易見到

更新3:

明左.. thx 幫埋我呢條啊, thx

回答 (3)

2012-03-07 8:07 am
✔ 最佳答案
(a) a + b + c = 0 => a = -(b + c)

a^3 + b^3 + c^3 + 3abc

= -(b + c)^3 + b^3 + c^3 - 3bc(b + c)

= -(b^3 + 3bc(b + c) + c^3) + b^3 + c^3 - 3bc(b + c)

= 0

(b) Consider a = x - 2√2 + 5√3, b = -3x - 6√2 + 7√3, c = 2x + 8√2 - 12√3a + b + c

= x - 2√2 + 5√3 -3x - 6√2 + 7√3 + 2x + 8√2 - 12√3

= -2√2 + 5√3 - 6√2 + 7√3 + 8√2 - 12√3

= 0

So, the original equation become

3abc = 0

3(x - 2√2 + 5√3)(-3x - 6√2 + 7√3)(2x + 8√2 - 12√3) = 0

x = 2√2 - 5√3, (7√3 - 6√2)/3, 6√3 - 4√2
2012-03-09 3:02 am
thz Kenneth
2012-03-07 8:22 am
[8^(1/3) = 2]

c^3 = 8(x + 4√2 - 6√3)^3
= 2^3 x ( x + 4√2 - 6√3 )^3
= [ 2 ( x + 4√2 - 6√3 ) ] ^3 <---- law of index, a^m x b^m = (ab)^m
= (2x + 8√2 - 12√3) ^3


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