Question

1/X-1-1/X+1-1/X²+1-1/X四次方+1=2X/X八次方-1

Answer 1

1/(X-1)-1/(X+1)-1/(X²+1)-1/(X^4+1)=2X/(X^8-1)

可知x≠1且x≠-1.

1/(X-1)-1/(X+1)=[(X+1)-(X-1)]/(X-1)(X+1)=2/(x^2-1)

2/(x^2-1)-1/(X²+1)=(x^2+3)/(x^4-1)

(x^2+3)/(x^4-1)-1/(X^4+1)

=[(x^2+3)*(x^4+1)-(X^4-1)]/[(x^4-1)(X^4+1)]

=[(x^2+2)*(x^4+1)+2]/[(x^8-1)]

所以原式可变为

(x^2+2)*(x^4+1)+2=2x

x^6+2x^4+x^2-2x+4=0

因为,在实数范围内

x^6+2x^4+x^2-2x+4>x^2-2x+4=(x-1)^2+3>0恒成立

所以方程无实数解.