Question

已知a^2-3a+1=0,求代数式a^2/a^4+1的值。快!!!详细

Answer 1

a^2-3a+1=0两边同除a,得a-3+1/a=0,即a+1/a=3,两枯州边圆败培橘唯平方得a^2+1/a^2+2=9,即a^2+1/a^2=7.
因此a^2/(a^4+1)=1/(a^2+1/a^2) = 1/7

Answer 2

a^2-3a+1=0
a^2+1=3a
a+1/隐银旁a=3
(a+1/搏晌a)^2=3^2
a^2+2+1/a^2=9
a^2+1/a^2=7
(a^4+1)/灶橡a^2=7
a^2/(a^4+1)=1/7

Answer 3

a^2-3a+1=0可以求出a=(3±√5)/2
则a^2=(7+3√5)/2或(7-3√唤梁5)/2
所以a^2/a^4+1=(18-3√5)/4或和凯运孙返(18+3√5)/4

Answer 4

a^2-3a+1=0
a^2+1=3a
a+1/隐银旁a=3
(a+1/搏晌a)^2=3^2
a^2+2+1/a^2=9
a^2+1/a^2=7
(a^4+1)/灶橡a^2=7
a^2/(a^4+1)=1/7