已知x+x分之1=3,求x^4+x^2+1分之x^2的值

2025-01-07 19:22:57
推荐回答(2个)
回答1:

x^2/(x^4+x^2+1) 取倒数

(x^4+x^2+1)/x^2
=x^2+1+1/x^2
=x^2+2+1/x^2-1
=(x+1/x)^2-1
=3^2-1
=8 (取倒数)

所以x^2/(x^4+x^2+1)=1/8

x/(x^2-3x+1)=7 取倒数
(x^2-3x+1)/x=1/7
x-3+1/x=1/7
x+1/x=22/7

x^2/(x^4+x^2+1) 取倒数

(x^4+x^2+1)/x^2
=x^2+1+1/x^2
=x^2+2+1/x^2-1
=(x+1/x)^2-1
=(22/7)^2-1
=435/49 (取倒数)

所以x^2/(x^4+x^2+1)=49/435

回答2:

都是把分子或者分母同除以某个式子的类型。