已知x^2-x-1=0,求x^4+1⼀2x^2的值. 大哥大姐帮帮忙啦啦啦

2024-11-23 12:52:07
推荐回答(3个)
回答1:

x^2-x-1=0
x²=x+1
所以x^4+x^2/2=(x+1)散兄²+(x+1)/2
令t=x+1
因为x²=x+1,所以t>0
所以x^4+1/2x^2=t²+t/2=(t+1/4)²-1/16
所握掘销以值为(段游0,+无穷)

回答2:

x²-x=1,亩首x²=x+1
x^4+x²/2=x²(2x²+1)/2
=(x+1)[2(x+1)+1]/2
=(x+1)(2x+3)/2
=(2x²+5x+3)/2
=[2(x+1)+5x+3]/2
=(7x+5)/2.
将乱历x=(1±√5)/2代入哗耐搜即可.

回答3:

x^2-x-1=0
∴x-1-1/x=0
x-1/x=1

(x^4+1)/数腊岁局猜2x^2
=1/2(x² +1/x²)
=1/2[(x-1/薯睁x)²+2]
=1/2(1²+2)
=3/2