MATHS 3條(函數) 10點

1) a= √5+1
a+1/a = (√5+1) + 1/(√5+1)
= √5+1+ {1/(√5+1) * [(√5-1) / (√5-1)]}
= √5+1 + (√5-1)/ [(√5)^2-1]
= √5+1 + (√5-1)/4
= (5√5+3)/4

(a+1/a)^2 = a^2 + 2*a* 1/a + 1/a^2
(a+1/a)^2 = a^2 + 2 + 1/a^2

a^2 + 1/a^2 = (a+1/a)^2 - 2
= [(5√5+3)/4]^2 - 2
= (125+30√5+9)/16 - 2
= (125+30√5+9-32)/16
= (102+30√5)/16
= (51+15√5)/8



3) If the above is wrong ans, i think it s/b f(x)=1/(x+1)及g(x)=1/x. If so...
f[g(x)] = 1/ [g(x)+1]
= 1/ [1/x+1]
= 1/ [(1+x)/x]
= x/ (1+x) OR 1+1/x

1.已知a=√5+1

a)求a+1/a的值
a+1/a
=(a^2+1)/a
=(5+2√5+1+1)/(√5+1)
=(7+2√5)/(√5+1)

b)藉展開(a+1/a)² ,計算a² + 1/a²
(a+1/a)²=a²+2+1/a²
a² + 1/a²
=(a+1/a)²-2
=(7+2√5)²/(√5+1)²-2
=(69+28√5)/(6+2√5)-2


2.已知f(x)=k²x-7
a)若f(k)=1,求k的值
f(k)=1
k3-7=1
k3=8
k=2


b)根據a小題的結果,解方程f(x)=0
k²x-7=0
4x=7
x=7/4

3.已知f(x)=1/x+1及g(x)=1/x

求f[g(x)]
f[g(x)]
=1/(1/x)+1
=1+x