假设(a√2+b)/(b√2+c)=k(有理数)
(a-bk)√2=kc-b(有理数)
所以:a-bk=0,kc-b=0
a=bk,c=b/k
(a^2+b^2+c^2)/(a+b+c)
=(b^2k^2+b^2+b^2/k^2)/(bk+b+b/k)
=b(k^4+k^2+1)/[k(k^2+k+1)]
=c[k^4+k^3+k^2-(k^3-1)]/(k^2+k+1)
=c[(k^2-k+1)(k^2+k+1)/(k^2+k+1)
=c(k^2-k+1)
假设(a√2+b)/(b√2+c)=k(有理数)
(a-bk)√2=kc-b(有理数)
所以:a-bk=0,kc-b=0
a=bk,c=b/k
(a^2+b^2+c^2)/(a+b+c)
=(b^2k^2+b^2+b^2/k^2)/(bk+b+b/k)
=b(k^4+k^2+1)/[k(k^2+k+1)]
=c[k^4+k^3+k^2-(k^3-1)]/(k^2+k+1)
=c[(k^2-k+1)(k^2+k+1)/(k^2+k+1)
=c(k^2-k+1)