✔ 最佳答案
1.
設k=log_9 (a)=log_12(b)=log_16(a+b), then
a=9^k, b=12^k, a+b=16^k
9^k*16^k=144^k, thus, a(a+b)=b^2
b^2-ab-a^2=0, t^2-t-1=0, (t=b/a)
t=(1+/- √5)/2 (負不合)
2.
z(z^4+1)=1, |z||z^4+1|=1, |z^4+1|=1
Let z=cosx+isinx, 0<x<2π, then
(cos4x+1)^2+(sin4x)^2=1
2+2cos(4x)=1, cos(4x)=-1/2,
4x= 2π/3, 4π/3, 8π/3, 10π/3, 14π/3, 16π/3, 20π/3, 22π/3
x=π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6
代回z(z^4+1)=1, then
x=π/3, 5π/3, so, z=cosx+isinx= (1 +/- i√3)/2
3.
設L: y-2=m(x-1)
與xy=1聯立,then 1/x -2=m(x-1), mx^2-(m-2)x-1=0
D=(m-2)^2+4m
(x1-x2)^2=(x1+x2)^2-4x1x2=[(m-2)/m]^2+4/m=1+ 4/m^2
(y1-y2)^2=m^2(x1-x2)^2
so, PQ^2=(m^2+1)(1+4/m^2)=5+m^2+ 4/m^2
>=5+4=9
min PQ=3 (m=2, -2)
Note: D=(m-2)^2+4m>0, m=2,-2均可
2010-06-30 22:37:52 補充:
哇!老王的方法真好:
|z^5|=|z|^5=|z-1|=1, 故找|z|=1 and |z-1|=1的交點即可
so, z=(1+/-√3)/2, 再代回z^5+z-1=0檢驗即可
2010-07-01 12:24:16 補充:
PQ^2=(m^2+1)(1+4/m^2)=5+m^2+ 4/m^2 >=5+4=9
min PQ=3 (m=√2, -√2) (Thanks!)
Note: D=(m-2)^2+4m=m^2+4>0, m=√2,-√2均可