已知x+1/x=2 求(1)x^2+1/x^2 (2)x^4+1/x^4

x+1/x=2 => x^2+2x+1=0 => x=1

x^2+1/x^2=2

x^4+1/x^4=2

x=1,后面自己算

x+ 1/x=2 两边同时乘以x
x^2+1-2*x=0 (x-1)=0 x=1

(1)1^2+1=2
(2)1+1=2

x^2+1/x^2 =[x+(1/x)]^2 = 4-2=2
x^4+1/x^4 = [x^2+1/(x^2)]^2 = 4-2=2

( x + 1/x )^2 = x^2+1/x^2 +2 = 4 ===> x^2+1/x^2 =2

(x^2+1/x^2)^2 = x^4+1/x^4 + 2 = 4 ====> x^4+1/x^4 = 2

另一方面：由 x+1/x=2 ==> x=1 ==> x^n + 1/x^n = 2

因为x+1/x=2 所以(x + 1/x )^2=4
(x + 1/x )^2 = x^2+1/x^2 +2
所以 x^2+1/x^2 =2
同理(x^2+1/x^2)^2 = x^4+1/x^4 +2
x^4+1/x^4 = 2