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Pe: 2012-03-24T08:19:21+02:00 In: In:

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Determinati numerele x, y, z stiind ca sunt direct proportionale cu nr 0,25 . 0,(3) . 0,5 si ca xy+xz+yz=3xyz

Am incercat de mai multe ori dar imi da niste rezultate absurde.

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2 raspunsuri

  1. bedrix
    expert (VI)

    x/0,25 = y/0,(3)=z/0,5 sau 4x=3y=2z |: 12 ; [4,3,2]=12
    x/3=y/4=z/6=k rezulta x=3k ; y=4k ; z=6k

    xy+xz+yz=3xyz echivalent cu 12k^2 +18k^2 + 24k^2=3*72k^3 rezulta k=1/4 rezulta cerinta

  2. doctal

    \rm{ \mathit {x}, y, z sunt direct proportionale cu 0,25 ; 0,(3) ; 0,5. (1)\\\;\\0,25 = \frac{1}{4} ; 0,(3) = \frac{3}{9} = \frac{1}{3} ; 0,5 = \frac{1}{2} (2)\\\;\\(1), (2) \Longrightarrow \frac{\mathit x}{\frac{1}{4}} = \frac{y}{\frac{1}{3}} = \frac{z}{\frac{1}{2}} \Longrightarrow 4\mathit x = 3y = 2z \Longrightarrow \{x = \frac{3y}{4}\\\;\\ \ \ \ \ \ \ (3)\\\;\\z = \frac{3y}{2} }

    \it{xy+yz+zx = 3xyz \stackrel{(3)}\Longrightarrow y =1 (4)\\\;\\(3), (4) \Longrightarrow x = \frac{3}{4} ; z = \frac{3}{2}.}

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