已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/08 20:44:10
![已知x/x²+x+1=a,试求分式x²/x四次方+x²+1](/uploads/image/z/469514-2-4.jpg?t=%E5%B7%B2%E7%9F%A5x%2Fx%26%23178%3B%2Bx%2B1%3Da%2C%E8%AF%95%E6%B1%82%E5%88%86%E5%BC%8Fx%26%23178%3B%2Fx%E5%9B%9B%E6%AC%A1%E6%96%B9%2Bx%26%23178%3B%2B1)
x){}K++Ԕ
-+
mu^lcӎ{2Og~fi;a
m)5BΆ*@ҭiEӯ5O2+`@l.OZ8#6 gk,gaki_\g
0-Z@%Ɓ@b`Ov/~ں|F0H. A
已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
已知x/x²+x+1=a,试求分式x²/x四次方+x²+1
x/(x²+x+1)=a
(x²+x+1)/x=1/a=x+1/x+1
x+1/x=1/a-1
x²/x四次方+x²+1
=1/(x^2+1/x^2+1)
=1/[(x+1/x)^2-1]
=1/[(1/a-1)^2-1]
=a^2/(1-2a)
x/x²+x+1=a
1/x+x=a-1
x²/x四次方+x²+1=1/x²+x²+1=(1/x+x)²-2+1
代入
(a-1)²-1=a²-2a