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andrei_g
Pe: 2013-07-30T20:18:42+03:00 In: In:

algebra

1. Sa se demonstreze ca:

    \[\sqrt1+\sqrt2+...+\sqrt{215}>2005\]

2. Sa se rezolve ecuatiile si sistemele de ecuatii:

    \[\begin{array}{l} a)\;\left\{ \begin{array}{l} \left[ x \right] + 2\left[ y \right] = 4\\ {x^2} + {y^2} = \frac{{17}}{2}\quad ;x > 0,y > 0\;si\;x + y \in N \end{array} \right.\\ b)\;\left[ x \right] \cdot \left\{ x \right\} = 1 \end{array}\]

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6 raspunsuri

  1. bedrix
    expert (VI)

    1) SQRT1=1
    2>SQRT2>1
    2>SQRT3>1
    SQRT4=2
    3>SQRT5>2
    .
    .
    .
    SQRT196=14
    15>SQRT197>14
    .
    .
    .
    15>SQRT215>14 ; insumam si obtinem :
    (SQRT1 + SQRT2 + SQRT3) + (SQRT4 + SQRT5 +…+SQRT8)+…+(SQRT196+…+SQRT215) > 3*1 + 5*2+…+20*14=…

  2. gunty
    maestru (V)

    2. – a)

        \[ \left\{ \begin{array}{l} \left[ x \right] + 2\left[ y \right] = 4 \\ x^2 + y^2 = \frac{{17}}{2} \\ \end{array} \right.;x,y > 0,x + y \in N \Leftrightarrow \left\{ \begin{array}{l} \left[ x \right] + 2\left[ y \right] = 4 \\ x + y = \sqrt {\frac{{17 + 4xy}}{2}} \in N \Rightarrow 17 + 4xy = 2k^2 ,k \in N \\ \end{array} \right. \]

        \[ \begin{array}{l} I.\;Daca\;\left[ y \right] = 1: \\ \Rightarrow \left[ x \right] = 2 \Rightarrow \\ \Rightarrow x \in \left[ {2,3} \right),y \in \left[ {1,2} \right) \Rightarrow xy\left[ {2,6} \right) \Leftrightarrow 17 + 4xy \in \left[ {25,41} \right) \\ \left. \begin{array}{l} 17 + 4xy \in \left[ {25,41} \right) \\ 17 + 4xy = 2k^2 \\ \end{array} \right\} \Rightarrow 17 + 4xy = 32 \Leftrightarrow x = \frac{{15}}{{4y}}. \Rightarrow \frac{{15}}{{4y}} + y = 4 \Leftrightarrow 4y^2 - 16y + 15 = 0 \Rightarrow \left\{ \begin{array}{l} y_1 = \frac{5}{2}\; \\ y_2 = \frac{3}{2} \\ \end{array} \right. \\ Deoarece\;y_1 \;nu\;respecta\;conditia\;\left[ y \right] = 1 \Rightarrow y = \frac{3}{2} \Rightarrow x = \frac{5}{2}. \\ \end{array} \]

        \[ \begin{array}{l} II.\;Daca\;\left[ y \right] \ge 2: \\ \Rightarrow \left[ x \right] \le 0 \Leftrightarrow x \le 0 \Rightarrow nu\;avem\;solutii. \\ \end{array} \]

        \[ In\;concluzie,\;\left( {x,y} \right) = \left( {\frac{5}{2},\frac{3}{2}} \right). \]

    2. – b)

        \[ \begin{array}{l} \left[ x \right]\left\{ x \right\} = 1\;,x \in R. \\ \\ Daca\;x \le 0 \Rightarrow \left[ x \right]\left\{ x \right\} \le 0 \Rightarrow nu\;avem\;solutii \\ Daca\;x \in \left( {0,1} \right) \Rightarrow 0 = 1_{\left( {fals} \right)} \\ Daca\;x \in \left[ {1,2} \right) \Rightarrow \left\{ x \right\} = 1_{\left( {imposibil} \right)} \\ Daca\;x \in \left[ {a,a + 1} \right)\;;a \in N,a \ge 2 \Rightarrow \left[ x \right] = a \Rightarrow \left\{ x \right\} = \frac{1}{a} \Rightarrow x = a + \frac{1}{a}. \\ \\ In\;concluzie,\;solutiile\;sunt\;x - urile\;de\;forma\;a + \frac{1}{a}\;,unde\;a \ge 2,a \in N. \\ \end{array} \]

  4. andu_flavius95
    maestru (V)

    gunty wrote: Daca\;x \le 0 \Rightarrow \left[ x \right]\left\{ x \right\} \le 0 \Rightarrow nu\;avem\;solutii \\

    Pai si daca x=-13/4 sau x=-3/2 , tot nu avem solutii reale ? 🙂

  5. gunty
    maestru (V)

    andu_flavius95 wrote: [quote=gunty]Daca\;x \le 0 \Rightarrow \left[ x \right]\left\{ x \right\} \le 0 \Rightarrow nu\;avem\;solutii \\

    Pai si daca x=-13/4 sau x=-3/2 , tot nu avem solutii reale ? 🙂

        \[ \left. \begin{array}{l} Daca\;x \le 0 \Rightarrow \left[ x \right] \le 0 \\ \left\{ x \right\} \in \left[ {0,1} \right)\;,\forall x \in R \\ \end{array} \right\} \Rightarrow \left[ x \right]\left\{ x \right\} \le 0 \Rightarrow \left[ x \right]\left\{ x \right\} \ne 1. \]

  6. andu_flavius95
    maestru (V)

    Asa este . Ai dreptate ,foarte precisa demonstratia . 🙂
    Scuze de interventie , din neatentie probabil am scris . Mi se mai intampla si mie . 😆

  8. gunty
    maestru (V)

    Am inteles… se mai intampla 😀

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