Question

已知a²+a-1=0,求值：a^4+1/a^ a+1/a

Answer 1

解：
∵a²+a-1=0
∴a≠0,两边除以a
得:a+1-1/a=0,即a-1/a=1
∴(a-1/a)²=1
a^2+1/a^2-2=1
a^2+1/a^2=3
(a+1/a)²=a^2+1/a^2+2=5
∴a+1/a=根号5
(a^2+1/a^2)^2=9
a^4+1/a^4+2=9
a^4+1/a^4=7

Answer 2

(1)
for all x ∈ A
=> x = f(x)
=>f(x) = f(f(x))
=>x = f(f(x)) ( x= f(x)
=> x ∈ B

then A is subset of B

(2)
f(x)=x^2+ax+b = x
x^2 + (a-1)x + b =0
sum of roots = -(a-1)/2 = 0
a= 1
products of roots = b = -4

ie f(x)= x^2 +x -4
for B = {x│x=f[f(x)]}
f(x) = x^2 +x - 4
f(f(x) )= ( x^2+x -4 )^2 +( x^2+x -4 )-4
x= ( x^2+x -4 )^2 +( x^2+x -4 )-4
x= (x^2-4)^2 + 2x(x^2-4) + x^2
+( x^2+x -4 )-4
x= x^4- 8x^2 +16 + 2x^3-8x + x^2
+( x^2+x -4 )-4
x^4+ 2x^3- 6x^2-8x+8 =0
(x^4-6x^2+8) + 2x(x^2-4)=0
(x^2-4)(x^2-2) + 2x(x^2-4)=0
(x^2-4)(x^2+2x-2)=0
x = 2 or -2 or -1+√3 or -1-√3

B={-1+√3 , -1-√3, 2, -2}

追问

啥意思呀，我的题给个答案都行

Answer 3

a²+a-1=0, a-1/a=-1
(a^4+1)/a^2=a^2+1/a^2=(a-1/a)^2+2=3

(a+1/a)^2=a^2+1/a^2+2=3+2=5
a+1/a= √5 a+1/a=-√5