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shervlad
Pe: 2012-12-28T09:47:56+02:00 In: In:

Ecuatie Exponentiala

Salutare!
Iata Ecuatia:
{ {x^2} \cdot {3^{x - 2}} + {3^{\sqrt x + 2}} = {3^x} + {x^2} \cdot {3^{\sqrt x }} \cr {x^2} \cdot {3^{x - 2}} + {3^{\sqrt x + 2}} = {x^2} \cdot {3^{\sqrt x }} + {3^x} \cr {x^2} \cdot {3^{x - 2}} - {3^x} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x + 2}} \cr {x^2} \cdot {{{3^x}} \over {{3^2}}} - {{{3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x + 2}} \cr {x^2} \cdot {{{3^x}} \over {{3^2}}} - {{{3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x }} \cdot {3^2} \cr {{{x^2}{3^x} - {3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) \cr {{{x^2}{3^x}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) \cr {{{3^x}} \over {{3^2}}} = {3^{\sqrt x }} \Leftrightarrow {3^{x - 2}} = {3^{\sqrt x }} \Leftrightarrow x - 2 = \sqrt x \Leftrightarrow \cr {x^2} - 4x + 4 = x \Leftrightarrow {x^2} - 5x + 4 = 0 \Leftrightarrow \cr {x_{1,2}} = {{5 \pm 3} \over 2} \cr {x_1} = 4 \cr {x_2} = 1 \cr}

problema e că la răspunsuri (la sfârsitul cărtii se dau solutiile 4 si 3) si când le înlocuiesc în ecuatie se primeste corect, iar 1 e un fel de „semisolutie” întrucât este corect doar pentru

    $$\sqrt 1 = - 1$$

când verific, parcă nu am făcut nici o eroare de calcul, as fi foarte recunoscător dacă m-ati ajuta…

Similare

2 raspunsuri

  1. Zeus
    veteran (III)

    Ai impartit printr-o relatie care poate fi 0 de-aia ai pierdut solutii.

  2. shervlad

    Mersi mult!!
    Iata si rezolvarea dupa parerea mea corecta, daca ii este cuiva interesant 🙂

        $${ {x^2} \cdot {3^{x - 2}} + {3^{\sqrt x + 2}} = {3^x} + {x^2} \cdot {3^{\sqrt x }} \cr {x^2} \cdot {3^{x - 2}} + {3^{\sqrt x + 2}} = {x^2} \cdot {3^{\sqrt x }} + {3^x} \cr {x^2} \cdot {3^{x - 2}} - {3^x} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x + 2}} \cr {x^2} \cdot {{{3^x}} \over {{3^2}}} - {{{3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x + 2}} \cr & {x^2} \cdot {{{3^x}} \over {{3^2}}} - {{{3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }} - {3^{\sqrt x }} \cdot {3^2} \cr {{{x^2}{3^x} - {3^x} \cdot {3^2}} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) \cr {{{x^2}{3^x}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} = {x^2} \cdot {3^{\sqrt x }}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) \cr {{{x^2}{3^x}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} - {x^2} \cdot {3^{\sqrt x }}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) = 0 \cr {{{x^2}{3^x}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} - {{{x^2} \cdot {3^{\sqrt x }} \cdot {3^2}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} = 0 \cr {{{x^2}{3^x}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) - {x^2} \cdot {3^{\sqrt x }} \cdot {3^2}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)} \over {{3^2}}} = 0 \cr {{{x^2}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)\left( {{3^x} - {3^{\sqrt x }}{3^2}} \right)} \over {{3^2}}} = 0 \cr {x^2}\left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right)\left( {{3^x} - {3^{\sqrt x }}{3^2}} \right) = 0 \cr \left[ \matrix{ \left( {1 - {1 \over {{x^2}}} \cdot {3^2}} \right) = 0 \Leftrightarrow 1 - {{{3^2}} \over {{x^2}}} = 0 \Leftrightarrow {{{3^2}} \over {{x^2}}} = 1 \Leftrightarrow {3^2} = {x^2} \Leftrightarrow x = 3 \hfill \cr {3^x} - {3^{\sqrt x }}{3^2} = 0 \Leftrightarrow {3^x} = {3^{\sqrt x }}{3^2} \Leftrightarrow x = \sqrt x + 2 \Leftrightarrow \hfill \cr} \right. \cr \Leftrightarrow x - 2 = \sqrt x \Leftrightarrow {x^2} - 4x + 4 = x \Leftrightarrow {x^2} - 5x + 4 = 0 \Leftrightarrow \cr {x_{1,2}} = {{5 \pm 3} \over 2} \cr {x_1} = 4 \cr {x_2} = 1,\,\,\,\sqrt 1 = - 1 \cr {x_3} = 3 \cr} $$

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