數學題目想破頭,想不出

Question

abc 不等於 0
a/b = b/c = c/a
那 (3a-2b+c) / (3a+2b-c) = ?
請大家幫個忙

謝謝你們的回答

不過這個問題是國中的題目

請用國中生能理解的方式來解決

那個JJ你解題是很有深度啦

深到見不到底

Answer 1

他是使用複數的觀點去探討

不過國中只講到實數
所以　你只要看他前面的1/2就好

Answer 2

題目本身只有一個限制abc≠0
所以將a,b,c各代1即可得到答案

Answer 3

a/b = b/c = c/a = k
=> a = bk, b = ck, c = ak
=> a = bk = ck2 = ak3
=> a = ak3
因為 abc 不等於 0
k3 = 1, => k = 1, ω, ω2
[ω= (-1+√3 i)/2, ω2 = (-1-√3 i)/2, ω2 + ω + 1 = 0]
k = 1 => a = b = c
(3a-2b+c) / (3a+2b-c) = (3a-2a+a) / (3a+2a-a) = 1/2
k = ω => a = bω, c = b/ω= bω2
(3a-2b+c) / (3a+2b-c)
= (3bω-2b+bω2) / (3bω+2b-bω2)
= (ω2+3ω-2) / (-ω2+3ω+2)
= (ω2+ω+1+2ω-3) / (-ω2-ω-1+ 4ω+3)
= (2ω-3)/(4ω+3)
= (-4+√3 i)/(1+2√3 i)
= (-4+√3 i)(1-2√3 i)/(1+13)
= (2+9√3 i)/14
k = ω2 => a = bω2, c = b/ω2= bω
(3a-2b+c) / (3a+2b-c)
= (3bω2-2b+bω) / (3bω2+2b-bω)
= (3ω2+ω-2) / (3ω2-ω+2)
= (3ω2+3ω+3-2ω-5) / (3ω2+3ω+3- 4ω-1)
= (-2ω-5)/(-4ω-1)
= (-3-√3 i)/(1-2√3 i)
= (-3-√3 i)(1+2√3 i)/(1+13)
= (3-7√3 i)/14
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.