急問:請問附圖的三題如何解呢？謝謝喔～?

(1)
a + 1/a = 4
a^2 - 4a + 1 = 0
a = [ 4 ± √(16-4) ] / 2 = 2 ± √3
a^(-1) = 1/a = 2 ± √3

令 b = a^(-1)
則 b - 2 = a^(-1) - 2 = ± √3

a^(-4) - 8*a^(-3) + 16*a^(-2) + 5
= b^4 - 8*b^3 + 16*b^2 + 5
= ( b^3 - 6b^2 + 4b + 8 )(b-2) + 21 , 利用綜合除法
= ( b^2 - 4b - 4 )(b-2)^2 + 21
= [ (b-2)^2 - 8 ](b-2)^2 + 21
= (b-2)^4 - 8(b-2)^2 + 21
= (±√3)^4 - 8(±√3)^2 + 21
= 9 - 24 + 21
= 6 ..... Ans

(2)
當 p + q = 1
f(p) + f(q)
= 16^p/( 4 + 16^p ) + 16^q/( 4 + 16^q )
= [ 16^p*( 4 + 16^q ) + 16^q*( 4 + 16^p ) ] / [ ( 4 + 16^p )( 4 + 16^q ) ]
= [ 4( 16^p + 16^q ) + 32 ] / [ 4( 16^p + 16^q ) + 32 ] , 因為 16^p*16^q = 16^(p+q) = 16^1 = 16
= 1

所以, 當 p + q = 1 , f(p) + f(q) = 1
1/6 + 5/6 = 1 , 故 f(1/6) + f(5/6) = 1
2/6 + 4/6 = 1 , 故 f(2/6) + f(4/6) = 1
3/6 + 3/6 = 1 , 故 f(3/6) + f(3/6) = 1 , 即 f(3/6) = 0.5
f(1/6) + f(2/6) + f(3/6) + f(4/6) + f(5/6) = 1 + 1 + 0.5 = 2.5 ..... Ans

(3)
0 = f(α) = [ ka^α - a^(-α) ] / [ a^α + a^(-α) ]
ka^α - a^(-α) = 0
ka^α = a^(-α)
a^(-2α) = a^(-α) / a^α = k
所以 a^(-2α) = k

當 f(x) = 1
a^x + a^(-x) = k*a^x - a^(-x)
2*a^(-x) = (k-1)a^x
a^(-2x) = a^(-x) / a^x = (k-1)/2

所以, 當 f(x) = 1 , a^(-2x) = (k-1)/2
又 f(β) = f(α+β) = 1
所以 a^(-2β) = a^(-2α-2β) = (k-1)/2

a^(-2α-2β) = a^(-2α) * a^(-2β)
(k-1)/2 = k * (k-1)/2
(k-1)(k-1)/2 = 0
k = 1 ..... Ans

1
a=p^2，b=q^2，p>0，q>0
b>a>0
q^2>p^2>0
q－p>0
[a－2√(ab)+b]^(1/2)－[a+2√(ab)+b]^(1/2)
=[p^2－2√(p^2q^2)+q^2]^(1/2)－[p^2+2√(p^2q^2)+q^2]^(1/2)
=(p^2－2pq+q^2)^(1/2)－(p^2+2pq+q^2)^(1/2)
=(q－p)－(p+q)
=－2p
=－2√a
2
f(2)=a－b*2^n=17
f(4)=a－b*4^n=77
f(8)=a－b*8^n=377
b*4^n－b*2^n=-60
b*8^n－b*4^n=-300
(4^n－2^n)/(8^n－4^n)=1/5
4^n+2^n=5
4^n+2^n－5>0
2^n>0
2^n=(-1+√21)/2 (負不合)
4^n=(1－2√21+21)/4=(11－√21)/2
b*(4^n－2^n)=-60
b*[(11－√21)/2－(－1+√21)/2]=-60
b*[(11－√21)－(－1+√21)]=－120
b*(12－2√21)=－120
b不為正整數
無解
3
Sol
設 a=10^667
[10^2001/(10^667+2002)]
=[a^3/(a+2002)]
=[(a^2－2002a+2002^2－2002^3/(a+2002))
=a^2－2002a+2002^2+[－2002^3/(a+2002)]
=a(a－2002)+2002^2+[－2002^3/(a+2002)]
=>2002^2－1
=>4008004－1
=>8003