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Alex2992
Pe: 2009-02-24T12:25:19+02:00 In: In:

sisteme si inecuatii cu logaritmi:D

am maine lucrare la matematica si incerc sa rezolv niste exercitii dar nu-mi ies toate
1

    \[ \begin{array}{l} rezolvati{\rm sistemul } \\ {\rm \{ 2}^{\rm x} + 2^y = 8 \\ {\rm log}_{\rm 2} x + \log _2 y = 2 \\ observam{\rm ca x = y = 2 este solutie si demonstram ca este unica? daca da}{\rm , cum facem asta?} \\ \end{array} \]

2

    \[ \begin{array}{l} {\rm rezolvati inecuatia log}_{\rm 3} (\frac{{x - 2}}{{x - 3}}) < 0 \\ il{\rm scriu pe 0 ca log}_{\rm 3} 1 = > (x - 2/x - 3) > 1? \\ \end{array} \]

3

    \[ {\rm rezolvati inecuatia }\frac{{{\rm x - 5}}}{{{\rm log}_{\rm 3} (x - 2)}} \ge 0 \]

4

    \[ \begin{array}{l} rezolvati{\rm sistemul } \\ {\rm \{ log}_{\rm 3} x + \log _3 y = 2 + \log _3 2 \\ {\rm log}{}_{\rm 3}(x + y) = 2 \\ aici{\rm nu am nicio idee} \\ \end{array} \]

Similare

4 raspunsuri

  1. ali
    maestru (V)

    rezolvare::

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 \left\{ \begin{array}{l}
	 \log _3 x + \log _3 y = \log _3 3^2  + \log _3 2 \\
	 \log _3 x + y = 2 \\
	 \end{array} \right. \Leftrightarrow  \\
	 \left\{ \begin{array}{l}
	 \log _3 xy = \log _3 18 \\
	 x + y = 9 \\
	 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
	 xy = 18 \\
	 x + y = 9 \\
	 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
	 P = 18 \\
	 S = 9 \\
	 \end{array} \right. \\
	 {\rm Formam ecuatia de gradul II in S si P ]
	

*** Error message:
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
Improper \prevdepth.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
Missing } inserted.
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Missing } inserted.
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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 10 ended by \end{document}.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Emergency stop.
  2. ali
    maestru (V)

    Alex2992 wrote: 3

        \[ {\rm rezolvati inecuatia }\frac{{{\rm x - 5}}}{{{\rm log}_{\rm 3} (x - 2)}} \ge 0 \]


    Pentru rezolvare se foloseste metoda standard cu „Tabel”:
    Rezolvam cele 2 inecuatii:

        \[ \begin{array}{l} 1)x - 5 \ge 0 \Rightarrow x \in [5,\infty ) \\ 2)\log _3 \left( {x - 2} \right) \ge 0 \Leftrightarrow \log _3 \left( {x - 2} \right) \ge \log _3 1 \Rightarrow x - 2 \ge 1 \Rightarrow x \in [3,\infty ) \\ 1) + 2) \Rightarrow x \in [5,\infty ). \\ \end{array} \]

  4. ali
    maestru (V)

    Alex2992 wrote: 2:

        \[ \begin{array}{l} {\rm rezolvati inecuatia log}_{\rm 3} (\frac{{x - 2}}{{x - 3}}) < 0 \\ il{\rm scriu pe 0 ca log}_{\rm 3} 1 = > (x - 2/x - 3) > 1? \\ \end{array} \]


    Da este corect insa:
    (x – 2/x – 3) < 1->(x – 2/x – 3)-1<0 apoi metoda „
    TABEL”

  5. ali
    maestru (V)

    Alex2992 wrote: 1

        \[ \begin{array}{l} rezolvati{\rm sistemul } \\ {\rm \{ 2}^{\rm x} + 2^y = 8 \\ {\rm log}_{\rm 2} x + \log _2 y = 2 \\ observam{\rm ca x = y = 2 este solutie si demonstram ca este unica? daca da}{\rm , cum facem asta?} \\ \end{array} \]


    Pentru demonstrare de obicei se demonstreaza ca functia este
    Bijectiva ceea ce duce la faptul ca functia are o solutie unica.
    Pentru a demonstra ca fucntia este bijectia este necesar ori:
    a)injectivitate+surjectivitate
    b)aratam ca functia este strict crescatoare\descrescatoare. 💡

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