Tìm Min D=2x^2+9y^2-6xy-6x+12y+2012 E=x^2-2xy+4y^2-2x+10y+29 F=3/(2x-x^2-4) G=2/(6x-5-9x^2)

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Tìm Min
D=2x^2+9y^2-6xy-6x+12y+2012
E=x^2-2xy+4y^2-2x+10y+29
F=3/(2x-x^2-4)
G=2/(6x-5-9x^2)

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Abigail 4 tháng 2021-08-11T22:55:09+00:00 1 Answers 5 views 0

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    0
    2021-08-11T22:56:50+00:00

    Giải thích các bước giải:

     Ta có:

    \(\begin{array}{l}
    *)\\
    D = 2{x^2} + 9{y^2} – 6xy – 6x + 12y + 2012\\
     = \left( {{x^2} – 6xy + 9{y^2}} \right) – 4.\left( {x – 3y} \right) + \left( {{x^2} – 2x + 1} \right) + 2011\\
     = {\left( {x – 3y} \right)^2} – 4.\left( {x – 3y} \right) + {\left( {x – 1} \right)^2} + 2011\\
     = \left[ {{{\left( {x – 3y} \right)}^2} – 4.\left( {x – 3y} \right) + 4} \right] + {\left( {x – 1} \right)^2} + 2007\\
     = {\left( {x – 3y – 2} \right)^2} + {\left( {x – 1} \right)^2} + 2007 \ge 2007,\,\,\forall x,y\\
     \Rightarrow {D_{\min }} = 2007 \Leftrightarrow \left\{ \begin{array}{l}
    {\left( {x – 3y – 2} \right)^2} = 0\\
    {\left( {x – 1} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x = 1\\
    y =  – \frac{1}{3}
    \end{array} \right.\\
    *)\\
    E = {x^2} – 2xy + 4{y^2} – 2x + 10y + 29\\
     = \left( {{x^2} – 2xy + {y^2}} \right) – 2.\left( {x – y} \right) + \left( {3{y^2} + 8y + \frac{{16}}{3}} \right) + \frac{{71}}{3}\\
     = {\left( {x – y} \right)^2} – 2\left( {x – y} \right) + 1 + 3.\left( {{y^2} + \frac{8}{3}y + \frac{{16}}{9}} \right) + \frac{{68}}{3}\\
     = {\left( {x – y} \right)^2} – 2.\left( {x – y} \right) + 1 + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3}\\
     = {\left( {x – y – 1} \right)^2} + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3} \ge \frac{{68}}{3},\,\,\,\forall x,y\\
     \Rightarrow {E_{\min }} = \frac{{68}}{3} \Leftrightarrow \left\{ \begin{array}{l}
    x – y – 1 = 0\\
    y + \frac{4}{3} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x =  – \frac{1}{3}\\
    y =  – \frac{4}{3}
    \end{array} \right.\\
    *)\\
    F = \frac{3}{{2x – {x^2} – 4}}\\
    2x – {x^2} – 4 =  – \left( {{x^2} – 2x + 1} \right) – 3 =  – {\left( {x – 1} \right)^2} – 3 \le  – 3,\,\,\,\forall x\\
     \Rightarrow F = \frac{3}{{ – {{\left( {x – 1} \right)}^2} – 3}} \ge \frac{3}{{ – 3}} =  – 1\\
     \Rightarrow {F_{\min }} =  – 1 \Leftrightarrow x = 1\\
    *)\\
    G = \frac{2}{{6x – 5 – 9{x^2}}}\\
    6x – 5 – 9{x^2} =  – \left( {9{x^2} – 6x + 1} \right) – 4 =  – 4 – {\left( {3x – 1} \right)^2} \le  – 4\\
     \Rightarrow G \ge \frac{2}{{ – 4}} =  – \frac{1}{2}\\
     \Rightarrow {G_{\min }} =  – \frac{1}{2} \Leftrightarrow x = \frac{1}{3}
    \end{array}\)

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