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？
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(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^10x/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=1x^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
======以下答案可供参考======
供参考答案1:
1024,注用平方差公式化简得x^10,