数学题:难题设2006x3=2007y3=2008z3,且xyz>0,3√2006x2+2007y2+2008z2 =3
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数学题:难题
设2006x3=2007y3=2008z3,且xyz>0,3√2006x2+2007y2+2008z2 =3√2006+3√2007+ 3√2008,求1/x+1/y+1/z
设2006x3=2007y3=2008z3,且xyz>0,3√2006x2+2007y2+2008z2 =3√2006+3√2007+ 3√2008,求1/x+1/y+1/z
设2006x^3=2007y^3=2008z^3=k
(2006x2+2007y2+2008z2)^(1/3)=2006^(1/3)+2007^(1/3)+2008^(1/3)
==>
(k/x+k/y+k/z)^(1/3)=(k/x^3)^(1/3)+(k/y^3)^(1/3)+(k/z^3)^(1/3)
==>k^(1/3)(1/x+1/y+1/z)^(1/3)=k^(1/3)(1/x+1/y+1/z)
==>1/x+1/y+1/z=(1/x+1/y+1/z)^(1/3)
==>1/x+1/y+1/z=1,或0(xyz>0,所以舍去),或-1(xyz>0,所以舍去)
故1/x+1/y+1/z=1
赠人玫瑰,手留余香!
(2006x2+2007y2+2008z2)^(1/3)=2006^(1/3)+2007^(1/3)+2008^(1/3)
==>
(k/x+k/y+k/z)^(1/3)=(k/x^3)^(1/3)+(k/y^3)^(1/3)+(k/z^3)^(1/3)
==>k^(1/3)(1/x+1/y+1/z)^(1/3)=k^(1/3)(1/x+1/y+1/z)
==>1/x+1/y+1/z=(1/x+1/y+1/z)^(1/3)
==>1/x+1/y+1/z=1,或0(xyz>0,所以舍去),或-1(xyz>0,所以舍去)
故1/x+1/y+1/z=1
赠人玫瑰,手留余香!
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