设x+1/x=-1,求下列各式的值:(1)x^3n+1/(x^3n);(2)x^(3n±1)+1/[x^(3n±1)];(3)当x变化时,求分式(3x^2+6x+5)/[(1/2)x^2+x+1]的最小值.

Question

设x+1/x=-1,求下列各式的值:
(1)x^3n+1/(x^3n);
(2)x^(3n±1)+1/[x^(3n±1)];
(3)当x变化时,求分式(3x^2+6x+5)/[(1/2)x^2+x+1]的最小值.

Answer 1

由x+1/x=-1
x^2+x+1=0
(x-1)(x^2+x+1)=0
得x^3-1=0
x^3=1,x1=-1/2+i根号3/2或x2 =-1/2-i根号3/2
（1）x^3n+1/(x^3n)=2
(2)x^(3n±1)+1/[x^(3n±1)]=x^(±1)+1/[x^(±1)]
x^(1)+1/[x^(1)]=1,x^(-1)+1/[x^(-1)]=x+1/x=1
(3)本问应当与前面 题目条件无关否则是复数谈不上最大最小.
y=(3x^2+6x+5)/[(1/2)x^2+x+1]
6x^2+12x+10=yx^2+2yx+2y
(6-y)x^2+(12-2y)x+10-2y=0
判别式=（12-2y)^2-8(6-y)(5-y)=144-48y+4y^2-240+88y-8y^2=-4y^2+40y-96>=0
y^2-10y+24