✔ 最佳答案
方程有好多解法,我可以有無限個方法..
方法1 - 你的方法:
5(n-3)=10
n-3 = 10/5
n-3 = 2
n = 2+3 = 5
方法2:
5(n-3) = 10
5(n) - 5(3) = 10 <--- 乘法分配性質
5n = 10 + 15
5n = 25
n = 25/5 = 5
以上方法用左expansion(f.2會學):a(b+c) = ab + ac
假設你唔明白5(n-3) = 5n - 5(3)....
當"5" 係一個蘋果...
n個蘋果減3個蘋果,等於(n-3)個蘋果
換句話,n(5) - 3(5) = (n-3)(5)...
依個應該明...
方法3:
5(n-3) = 10
5(n-3) - 10 = 0
5(n-3) - 5(2) = 0
5(n-3-2) = 0
∴n - 3 - 2 = 0
n = 3+2 = 5
如上,只不過係expansion 倒轉 factorization: ab + ac = a(b+c)
(n-3) 個蘋果 減 2個蘋果,就係(n-3-2) 個蘋果...
所以, (n-3)(5) - 2(5) = (n-3-2)(5)
另外,運用ab = 0 o既定理...
如果ab = 0,
a = 0 或 b = 0 都符合條方程。
如果已知其中一個數... 如
5a = 0
咁樣 a就一定等於0...
所以上面5(n-3-2) = 0
n - 3 - 2 = 0
甚至有更加無厘頭o既方法 (勿學,不過可以嘗試去明白):
方法4:
5(n-3) = 10
5(n-3)/10 = 1
0.5(n-3) = 1
0.5n - 1.5 = 1
0.5n - 1.5 - 1 = 0
0.5n - 2.5 = 0
0.5(n-5) = 0
n - 5 = 0
n = 5
方法5:
5(n-3) = 10
log[5(n-3)] = log10
log5 + log(n-3) = log10
log(n-3) = log10 - log5
log(n-3) = log(10/5)
n-3 = 2
n = 2+3 = 5
方法6:
5(n-3)=10
設a = 5 <--- 將所有已知都轉做純"a", 所以 3 = a-2 & 10 = 2a
a[n-(a-2)] = 2a
an - a(a-2) = 2a
an - a^2 + 2a = 2a
an - a^2 = 0
a(n-a) = 0
a = 0 (rej.) or n-a = 0
n = a = 5
駛唔駛再多d - -??
2010-01-31 22:03:12 補充:
講漏左個方法 - -... 將n由0,1,2,3.... 逐個估 xd... 都係一個方法~~