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cori
Pe: 2013-02-26T16:17:06+02:00 In: In:

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Sa se rezolve ecuatia: 3(n+1)!=(n+2)!

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7 raspunsuri

  1. xor_NTG
    maestru (V)

    Evident, n trebuie sa fie numar natural.

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	 3\left( {n + 1} \right)! = \left( {n + 2} \right)! \Leftrightarrow 3\left( {n + 1} \right)! = \left( {n + 1} \right)!\left( {n + 2} \right)\left| {]
	

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  2. gunty
    maestru (V)

        \[ \left( {n + 2} \right)! = \left( {n + 2} \right) \cdot \left( {n + 1} \right)! \]

    Deci egalitatea devine

        \[ 3 = n + 2 \Leftrightarrow n = 1 \]

    .

  4. cori

    Multumesc foarte mult.
    Daca ma puteti ajuta si la al doilea… nu le pot da de capat.
    (n+1)!(n+1)=6*(n+2) , n apartine la N

  5. gunty
    maestru (V)

        \[ \begin{array}{l} \left( {n + 1} \right)!\left( {n + 1} \right) = 6\left( {n + 2} \right) \\ \Leftrightarrow \left( {n + 1} \right)! = 6\left( {1 + \frac{1}{{n + 1}}} \right) \\ \end{array} \]

    Demonstram inegalitatea

        \[ 1 + \frac{1}{{n + 1}} \le 2 \]

    :

        \[ 1 + \frac{1}{{n + 1}} \le 2 \Leftrightarrow n + 1 + 1 \le 2\left( {n + 1} \right) \Leftrightarrow 1 \le n + 1_{\left( {adev.} \right)} \]

    . Deci avem

        \[ 6\left( {1 + \frac{1}{{n + 1}}} \right) \le 12 \]

    .
    In continuare, le luam pe cazuri:

        \[ \begin{array}{l} n = 0 \Rightarrow 1 = 6 \cdot 2_{\left( {fals} \right)} \\ n = 1 \Rightarrow 2 = 9_{\left( {fals} \right)} \\ n = 2 \Rightarrow 6 = 8_{\left( {fals} \right)} \\ n > 2 \Rightarrow \left( {n + 1} \right)! > 12 \Rightarrow \left( {n + 1} \right)! \ne 6\left( {1 + \frac{1}{{n + 1}}} \right) \\ \end{array} \]

    In concluzie, egalitatea nu are solutii naturale.

  6. cori

    Multumesc !

  8. cori

    Am mai revenit cu un set de exercitii…

  9. bedrix
    expert (VI)

    1. 5*n! +(n+1)!=5n*(n-1)! + (n+1)n*(n-1)!=40*(n-1)! | : (n-1)! devine 5n+n(n+1)=40 sau n^2 +6n -40 =0 sau (n+10)(n-4)=0 singura solutie este n=4

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