太多年沒有計Maths.很簡單的一條算式但已經唔記得點計. 請大家幫忙!
66 = (n²-n )/ 2
n = ??
請問呢條一元二次方程式點計?
2007-08-06 3:09 am
回答 (4)
2007-08-06 3:22 am
✔ 最佳答案
66 = (n²-n )/ 2n²-n=132
n²-n-132=0
132=2*66=2*2*33=2*2*3*11=11*12
so (n-12)(n+11)=0
n=12 or n=-11
2007-08-06 5:09 am
66 = (n²-n )/ 2
66x2=(n²-n)
n²-n-132=0
(n+11)(n-12)=0
i.e. (n+11)=0 or (n-12)=0
n= -11 or n=12
66x2=(n²-n)
n²-n-132=0
(n+11)(n-12)=0
i.e. (n+11)=0 or (n-12)=0
n= -11 or n=12
2007-08-06 3:15 am
66 = (n²-n )/ 2
132=(n²-n )
n²-n-132=0
(n+11)(n-12)=0
n+11=0 or n-12=0
n= -11 or n= 12
132=(n²-n )
n²-n-132=0
(n+11)(n-12)=0
n+11=0 or n-12=0
n= -11 or n= 12
2007-08-06 3:12 am
66 = (n²-n )/ 2
n²-n =132
n²-n -132=0
(n+11)(n-12)=0
n= 12 or -11
n²-n =132
n²-n -132=0
(n+11)(n-12)=0
n= 12 or -11
收錄日期: 2021-04-13 13:45:06
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