Question

已知a²－6a＋1=0,求（a-1／a)²的值分式

Answer 1

a²+1=6a
平方
a^4+2a²+1=36a²
a^4+1=34a²
两边除以a²
a²+1/a²=34

所以(a-1/a)²
=a²-2+1/a²
=34-2
=32

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}

Answer 3

a²－6a＋1=0
a-6+1/a=0
a+1/a=6
（a-1／a)²
=a^2+1/a^2-2
=(a+1/a)^2-4
=36-4
=32

Answer 4

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