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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Feb 2007 21:26:55 IST
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check ur question again i think that u have typed wrong i had solved this question i think the question is (1/ x3y3z3 ) ( x4y2z2+ 9x2y4z2 + 25 x2y2z4) = 15/x + 5/y + 3/z then it reduces to a2+b2+c2=ab+bc+ca where a=x2yz b=3xy2z c=5 xyz2 this expression again reduces to (a-b)^2+(b-c)^2+(c-a)^2=0 so a=b=c x=3y=5z no are x,x/3,x/5 so H.P
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18 Feb 2007 21:28:52 IST
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18 Feb 2007 21:30:51 IST
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Hi shakirshafi ! yes ! I totally agree with ur answer !!!! the question must be wrong !!!!!
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Umang |
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18 Feb 2007 21:42:30 IST
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SORRY !! I HAVE CORRECTED IT NOW ! GREAT MAN !!!
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18 Feb 2007 21:45:54 IST
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yeah its HP ...... shakir has done the right job (as he always do  )
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Manasi....
NIT-Allahabad...
............................................................
Challenges are High, Dreams r New..
The World out thr is waiting for U !!
Dare to dream, Dare to Try..
No Goal is distant, no Star is too high !!! |
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19 Feb 2007 11:21:14 IST
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Hey vish ! Let me remind you that this is the second question u hav posted incorrectly .It causes lot of problems .I hope next time there will be no problem !!!!!!! Keep posting good questions !!!
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Umang |
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19 Feb 2007 12:52:42 IST
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i m sorry !
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19 Feb 2007 16:51:53 IST
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question is wrong i feel but if made right the x,y,z are in hp as x=3y then and x=5z
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m |
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