1）因为 x^6+1-(x^4+x^2)
=x^4(x^2-1)-(x^2-1)
=(x^2-1)(x^4-1)
=[(x-1)(x+1)]^2(x^2+1)
当x=1或者-1时,x^6+1-(x^4+x^2)=0,即x^6+1=x^4+x^2
当x为其它任意数时,[(x-1)(x+1)]^2(x^2+1)0>,则 x^6+1>x^4+x^2
2）因为1+2x^4-2x^3-x^2
=2x^3(x-1)-(x-1)(x+1)
=(x-1)(2x^3-x-1)
=(x-1)[2(x^3-1)-(x-1)]
=(x-1)[2(x-1)(x^2+x+1)-(x-1)]
=(x-1)^2[2(x^2+x+1)-1]
=(x-1)^2(2x^2+2x+1)
又因为2x^2+2x+1=2（x^2+x）+1
=2[（x^2+x+1/4）-1/4]+1
=2[（x+1/2）^2-1/4]+1
=2（x+1/2）^2+1/2>0
因为 x≠1,则(x-1)^2>0
所以1+2x^4-2x^3-x^2>0
即1+2x^4>2x^3+x^2