ax=by=cz=1,求(1/1+a4)+(1/1+b4)+(1/1+c4)+(1/1+x4)+(1/1+y4)+(1/1+Z4)= (4:4次方)
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ax=by=cz=1,求(1/1+a4)+(1/1+b4)+(1/1+c4)+(1/1+x4)+(1/1+y4)+(1/1+Z4)=(4:4次方)ax=by=cz=1,求(1/1+a4)+(1/1+b
ax=by=cz=1,求(1/1+a4)+(1/1+b4)+(1/1+c4)+(1/1+x4)+(1/1+y4)+(1/1+Z4)= (4:4次方)
ax=by=cz=1,求(1/1+a4)+(1/1+b4)+(1/1+c4)+(1/1+x4)+(1/1+y4)+(1/1+Z4)= (4:4次方)
ax=by=cz=1,求(1/1+a4)+(1/1+b4)+(1/1+c4)+(1/1+x4)+(1/1+y4)+(1/1+Z4)= (4:4次方)
当ax=by=cz=1时,求1/(1+a^4) +1/(1+b^4) +1/(1+c^4) +1/(1+x^4) +1/(1+y^4) +1/(1+z^4)的值
是么?
a=1/x,b=1/y,c=1/z
所以1/(1+a^4)+1/(1+x^4)=1/(1+1/x^4)+1/(1+x^4)=x^4/(1+x^4)+1/(1+x^4)=1
其他四项两两一组,也可以求出来的
所以最后是
3