Question

已知ax=by=cz=1,求1+a的四次方分之1+1+b的四次方分之1+1+c的四次方分之1+1+x的四次方分之1+1+

Answer 1

1/(1+a^4)+1/(1+b^4)+1/(1+c^4)+1/(1+x^4)+1/(1+y^4)+1/(1+z^4)

=[1/(1+a^4)+1/(1+x^4)]+[1/(1+b^4)+1/(1+y^4)]+[1/(1+c^4)+1/(1+z^4]

∵ax=1

∴1/(1+a^4)+1/(1+x^4)

=x^4/[x^4+(ax)^4)+1/(1+x^4)]

=x^4/(x^4+1)+1/(1+x^4)

=(x^4+1)/(x^4+1)

=1

同理：1/(1+b^4)+1/(1+y^4)=1

1/(1+c^4)+1/(1+z^4=1

∴原式=1+1+1=3