設a,b,c都是正實數，若11,21,31是方程式a^(1/

log(abc) = 1

2013-01-18 19:55:20 補充：
Oh! 看錯，應該係：
log(abc) = 72

2013-01-19 11:37:49 補充：
令 a=10^p, b=10^q, c=10^r, 所以
a^(1/x) * b^[1/(x+3)] * c^[1/(x+6)] = 10
==> 10^(p/x) * 10^[q/(x+3)] * 10^[r/(x+6)] = 10^1
==> p/x + q/(x+3) + r/(x+6) = 1
==> px^2 + 9px + 18p + qx^2 + 6qx + rx^2 + 3rx = x^3 + 9x^2 + 18x
==> x^3 - (p + q + r)x^2 + (18 - 9p - 6q - 3r)x - 18p = 0

2013-01-19 11:42:41 補充：
對不起，最後一行是
x^3 - (p + q + r - 9)x^2 + (18 - 9p - 6q - 3r)x - 18p = 0

因為此式的根是 11, 21, 31, 所以
p + q + r - 9 = 11 + 21 + 31
==> p + q + r = 72
==> log(a) + log(b) + log(c) = 72
==> log(abc) = 72

螞蟻雄兵大

您第三式到第四式看來是等號兩邊都乘以 x(x+3)(x+6)

但是否右邊忘了乘？

解聯立方程式
答案72

設a，b，c都是正實數，若11，21，31是方程式a^(1/x)*b^[1/(x+3)]*c^[1/(x+6)]
=10的三個根，則log(abc)=?
Sol
a^(1/x)*b^[1/(x+3)]*c^[1/(x+6)]=10
log｛a^(1/x)*b^[1/(x+3)]*c^[1/(x+6)]｝=log10
(1/x)loga+[1/(x+3)]logb+[1/(x+6)]logc=1
(x+3)(x+6)loga+x(x+6)logb+x(x+3)logc=1
(x^2+9x+18)loga+(x^2+6x)logb+(x^2+3x)logc=1
(loga+logb+logc)x^2+(9loga+6logb+3logc)x+18loga-1=0
So
loga+logb+logc=0，9loga+6logb+3logc=0，18logc－1=0
log(abc)=0