Question

已知x／2＝y／3＝z／4,且xy－yz＋xz＝2.试求﹙x五次方＋y五次方﹚﹙x五次方-y五次方

Answer 1

(x^5+y^5)(x^5-y^5)+(y^5+z^5)(y^5-z^5)+(z^5)^2

=x^10-y^10+y^10-z^10+z^10

=x^10

x/2=y/3=z/4=a 则x=2a;y=3a;z=4a

6a^2-12a^2+8a^2=2

2a^2=2

a^2=1

x^10=(2a)^10=2^10

(x^5+y^5)(x^5-y^5)+(y^5+z^5)(y^5-z^5)+(z^5)^2=2^10