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Gabytza_u
Pe: 2013-08-06T18:52:34+03:00 In: In:

Buna

Am nevoie de ajutor la un exercitiu pe care nu l-am inteles niciodata.
M -ti putea ajuta,va rog 🙄 🙄 ❓ ❓
Cerinta exercitiului este : „Aflati:”

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3 raspunsuri

  1. bedrix
    expert (VI)

    c) 15/(4-k) E Z rezulta 4-k E D15{-15,-5,-3,-1,1,3,5,15} (1)
    k E N rezulta k >= 0 sau -k<=0 rezulta 4-k<=4 (2)
    Din (1) si (2) rezulta 4-k E {-15,-5,-3,-1,1,3}
    Analizezi fiecare caz
    Ex. pentru 4-k = -15 rezulta k=19 , continua tu

    d) 16/(2k+1) E Z rezulta 2k+1 E D16 (1)
    k E Z rezulta 2k+1 este impar (2)
    Din (1) si (2) rezulta 2k+1 E {-1,1} rezulta k

  2. Minecraft
    maestru (V) 
    *** QuickLaTeX cannot compile formula:

	\[\begin{array}{l}
	{\rm{Gabytza}}{\rm{, Bedrix a ales doua dintre cele 6 subpuncte ale exercitiului propus de tine}}{\rm{.}}\\
	{\rm{Daca recapitulezi putin teoria o sa te descurci si cu celelalte}}{\rm{.}}\\
	a){\rm{ Daca }}n \in {\rm{N si }}q \in {\rm{Q}}{\rm{, aflati valorile lui }}n{\rm{ pt care numarul rational }}q = \frac{{14}}{n} \in {\rm{N\backslash Q sau N - Q}}\\
	{\rm{Iata cateva din valorile lui }}q = \frac{{14}}{n}\\
	\frac{{14}}{0} = 0;\frac{{14}}{1} = 14;\frac{{14}}{{ - 1}} =  - 14;\frac{{14}}{2} = 7;\frac{{14}}{3} = 4,(6);etc.\\
	{\rm{Din conditiile puse le alegem doar pe acelea in care }}n \in {\rm{N}}{\rm{, adica }}n \in \left\{ {0,1,2,3,...} \right\}{\rm{ si }}q \in Z{\rm{, }}q \in \left\{ {0,1,2,3,...} \right\}\\
	14 = {2^1}*{7^1} \Rightarrow {\rm{Multimea divizorilor lui 14 are patru termeni]
	
	

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  4. Gabytza_u

    Minecraft wrote: 

    *** QuickLaTeX cannot compile formula:

	\[\begin{array}{l}
	{\rm{Gabytza}}{\rm{, Bedrix a ales doua dintre cele 6 subpuncte ale exercitiului propus de tine}}{\rm{.}}\\
	{\rm{Daca recapitulezi putin teoria o sa te descurci si cu celelalte}}{\rm{.}}\\
	a){\rm{ Daca }}n \in {\rm{N si }}q \in {\rm{Q}}{\rm{, aflati valorile lui }}n{\rm{ pt care numarul rational }}q = \frac{{14}}{n} \in {\rm{N\backslash Q sau N - Q}}\\
	{\rm{Iata cateva din valorile lui }}q = \frac{{14}}{n}\\
	\frac{{14}}{0} = 0;\frac{{14}}{1} = 14;\frac{{14}}{{ - 1}} =  - 14;\frac{{14}}{2} = 7;\frac{{14}}{3} = 4,(6);etc.\\
	{\rm{Din conditiile puse le alegem doar pe acelea in care }}n \in {\rm{N}}{\rm{, adica }}n \in \left\{ {0,1,2,3,...} \right\}{\rm{ si }}q \in Z{\rm{, }}q \in \left\{ {0,1,2,3,...} \right\}\\
	14 = {2^1}*{7^1} \Rightarrow {\rm{Multimea divizorilor lui 14 are patru termeni]
	
	

*** Error message:
\begin{array} on input line 9 ended by \end{document}.
leading text: \end{document}
Improper \prevdepth.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Missing \cr inserted.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
You can't use `\end' in internal vertical mode.
leading text: \end{document}
\begin{array} on input line 9 ended by \end{document}.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Emergency stop.

    Minecraft si Bedrix, m-ati ajutat foarte mult, si m-ati facut sa inteleg aceste exercitii.
    Va multumesc foarte mult!
    😛 😛 😛 😛 😀 😀

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