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Alesia96
Pe: 2013-04-09T12:53:07+03:00 In: In:

sa se rezolve in R si RxR

sa se rezolve in R si RxR

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2 raspunsuri

  1. gunty
    maestru (V)

    1.

        \[ \left( {\frac{x}{{x - 1}} + x - 2} \right)\left( {100^{\lg \left( {x - 1} \right)} - 2} \right) = 0,x > 1 \]

    Ai doua cazuri:

    I. Daca

        \[ \frac{x}{{x - 1}} + x - 2 = 0 \]

    :

        \[ \begin{array}{l} \Leftrightarrow x + \left( {x - 1} \right)\left( {x - 2} \right) = 0 \Leftrightarrow x^2 - 2x + 2 = 0 \\ \Delta = 4 - 8 = - 4 < 0 \\ \end{array} \]

    Deoarece delta<0, inseamna ca egalitatea nu are solutii reale.

    II. Daca

        \[ 100^{\lg \left( {x - 1} \right)} - 2 = 0 \]

    :

        \[ \Leftrightarrow \left( {10^{\lg \left( {x - 1} \right)} } \right)^2 = 2 \Leftrightarrow \left( {x - 1} \right)^2 = 2 \Rightarrow x - 1 = \pm \sqrt 2 \Leftrightarrow x = 1 \pm \sqrt 2 \]

    Deoarece avem conditia x>1, inseamna ca x=1-radical(2) nu e solutie.

    In concluzie, solutia egalitatii este x=1+radical(2).

    2.

    Se scriu conditiile de existenta: x-y>0 si x+y>0.

        \[ \begin{array}{l} \left\{ \begin{array}{l} 4\log _4 \left( {x - y} \right) + \log _2 \left( {x + y} \right) = 6 \\ \sqrt {x - y} = 2 \\ \end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l} 4\log _4 \left( {x - y} \right) + \log _2 \left( {x + y} \right) = 6 \\ x - y = 4 \\ \end{array} \right. \\ \end{array} \]

    Inlocuind x-y=4 in prima egalitate, primim:

        \[ 4\log _4 4 + \log _2 \left( {x + y} \right) = 6 \Leftrightarrow \log _2 \left( {x + y} \right) = 2 \Rightarrow x + y = 4 \]

    .

    Asadar sistemul devine:

        \[ \left\{ \begin{array}{l} x + y = 4 \\ x - y = 4 \\ \end{array} \right. \Rightarrow \left( {x,y} \right) = \left( {4,0} \right) \]

    .

  2. Alesia96

    ms mult

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