MQ64 --- Cubic Root
2012-12-31 7:37 pm
MQ64 --- Cubic RootDifficulty: 40% Given x = ∛4 + ∛2 + 1, evaluate (1 + 1/x)³.
回答 (3)
2012-12-31 8:04 pm
✔ 最佳答案
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2013-01-01 4:56 am
Both answers are good.
2012-12-31 9:30 pm
我有另一想法 又係啦 我第二位 唔洗比分我
LET y=∛2
x=y^2+y+1
So(1 + 1/x)³=(1+1/y^2+y+1)³=(1+y-1/(y^3-1))³=(1+[∛2 -1]/(2-1))³=[1+∛2 -1]³=(∛2)³
=2
我冇睇到...如果同第一位一樣 唔好扣我分T.T
2012-12-31 13:33:25 補充：
So(1 + 1/x)³=(1+1/(y^2+y+1))³=(1+(y-1)/(y^3-1))³=(1+[∛2 -1]/(2-1))³=[1+∛2 -1]³=(∛2)³=2
漏左( ) SOR
LET y=∛2
x=y^2+y+1
So(1 + 1/x)³=(1+1/y^2+y+1)³=(1+y-1/(y^3-1))³=(1+[∛2 -1]/(2-1))³=[1+∛2 -1]³=(∛2)³
=2
我冇睇到...如果同第一位一樣 唔好扣我分T.T
2012-12-31 13:33:25 補充：
So(1 + 1/x)³=(1+1/(y^2+y+1))³=(1+(y-1)/(y^3-1))³=(1+[∛2 -1]/(2-1))³=[1+∛2 -1]³=(∛2)³=2
漏左( ) SOR
參考: 自己
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