数学问题
已知X+X^-2=2,求(1)X^2+X^-2;(2)X^4/(X^8+X^4+1)
P.S.X^-2即X的负二次方,X^2,X^4……同理
参考答案:X+X^-2=2,
两边同时乘以X^2
X^3+1=2X^2
X^3-X^2-X^2+X-X+1=0
X^2(X-1)-X(X-1)-(X-1)=0
(X^2-X-1)(X-1)=0
∴X=1或X^2=X+1
(1)又可以从原式推出X^2=2-X
当X=1时
X^2+X^-2=1^2+1^(-2)=2
当X^2=X+1时
X^2+X^-2=X+1+2-X=3
所以X^2+X^-2=2或3
(2)X^4/(X^8+X^4+1)
当X=1时
X^4/(X^8+X^4+1)=1/(1+1+1)=1/3
当X^2=X+1时
X^4/(X^8+X^4+1)=(x+1)^2/((x+1)^4+(x+1)^2+1)
=(x^2+2x+1)/((x^2+2x+1)^2+(x^2+2x+1)+1)
=(3x+2)/((3x+2)^2+(3x+2)+1)
=(3x+2)/(9x^2+12x+4+3x+3)
=(3x+2)/(9(x+1)+15x+7)
=(3x+2)/(24x+16)
=1/8
所以X^4/(X^8+X^4+1)＝1/3或1/8