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shervlad
Pe: 2013-01-11T18:10:59+02:00 In: In:

Inecuatie exponentiala

{ {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr DVA:\,\,x + 2 \ge 0 \Leftrightarrow x \ge - 2 \cr {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {\left( {{1 \over 3}} \right)^x} \cr \sqrt {x + 2} > x \cr}

daca rudic ambele parti la patrat, imi da x e (-1;2) dar la solutii sa sfarsitul cartii este intervalul [-2;1), cum ajung la acea solutie?

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5 raspunsuri

  1. ex-admin
    profesor

    ridicarea la patrat poate fi facuta numai in situatia in care ambii membri sunt pozitivi.

    Cazul x<0 trebuie tratat separat.

  2. shervlad

    { {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr \sqrt {x + 2} > x \cr for\, - 2 \le x \le 0 \cr \sqrt {x + 2} > x \cr x \ge 0,\,\,x + 2 > {x^2} \cr {x^2} - x - 2 < 0 \cr x \in \left( { - 1;2} \right) \cr x \in \left[ { - 2;2} \right) \cr}
    asa o varianta este acceptabila?

  4. edy8
    maestru (V)

    shervlad wrote: { {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} }

    \it{\bl (\frac{1}{3})^{\sqrt{x+2}} < 3^{-x} \Leftrightarrow (\frac{1}{3})^{\sqrt{x+2}} < (\frac{1}{3})^x \Leftrightarrow \sqrt{x+2} > x (1)\\\;\\Rezolvam inecuatia (1)\\\;\\\sqrt{x+2} \Longrightarrow \{\sqrt{x+2} \geq0 (2)\\\;\\x+2\geq0 \Rightarrow x\geq -2 \Rightarrow x \in [ -2, \infty) \Rightarrow x \in [-2, 0) U [0, \infty ) (3)

    \it{\bl Se disting doua cazuri:\\\;\\I) x \in [-2, 0) \Rightarrow x<0 (4)\\\;\\(2), (4) \Rightarrow \sqrt{x+2} > x \forall x \in [-2, 0) \Rightarrow S_1 = [-2, 0)\\\;\\II) x \in [0, \infty) \Rightarrow inecuatia (1) devine:\\\;\\x+2 > x^2 \Leftrightarrow x^2 < x+2|_{ -1} \Leftrightarrow x^2-1 < x+1 \Leftrightarrow (x-1)(x+1) < x+1|_{ :(x+1)} \Leftrightarrow x-1 < 1 \Leftrightarrow x < 2 \Leftrightarrow \\\;\\\Leftrightarrow x \in [0, 2) \Rightarrow S_2 = [0, 2)\\\;\\Multimea solutiilor inecuatiei este :\\\;\\S = S_1 U S_2 = [-2, 0) U [0, 2) \Rightarrow S = [-2, 2)}

  5. shervlad

    Multumesc, dar se primeste ca ecuatia are solutii doar pentru \sqrt {x + 2} \ge 0
    ?
    pentur ca, sa presupunem x=-0.5 =>

  6. edy8
    maestru (V)

    shervlad wrote: Multumesc, dar se primeste ca ecuatia are solutii doar pentru \sqrt {x + 2} \ge 0
    ?
    pentur ca, sa presupunem x=-0.5 =>

    \it{\Large\bl \sqrt{1,5}\geq-0.5 \Longleftrightarrow 1,2 \geq- 0,5}

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