已知x/2=y/3=z/4,且xy-yz+xz=2,求(x^5+y^5)(x^5-y^5)+(y^5+z^5)(y^5-

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已知x/2=y/3=z/4,且xy-yz+xz=2,求(x^5+y^5)(x^5-y^5)+(y^5+z^5)(y^5-z^5)+(z^5)^2
从代数式的化简中，你发现了什么？

(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
再问：发现是什么？就是最后的式子吗
再答：发现:应用平方差能化简式子，减少大量计算

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