Tìm GTNN hoặc GTLN a) C = ( x^2 – 2x + 2021) / x^2 b) D = ( x^2 – 2x + 2015 ) / 2015x^2

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1. Đáp án:

    a) \(Min = \dfrac{{2020}}{{2021}}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)C = \dfrac{{{x^2} – 2x + 2021}}{{{x^2}}}\\
    = 1 – \dfrac{2}{x} + \dfrac{{2021}}{{{x^2}}}\\
   Đặt:\dfrac{1}{x} = t\\
    \to C = 1 – 2t + 2021{t^2}\\
    = 2021{t^2} – 2.t\sqrt {2021} .\dfrac{1}{{\sqrt {2021} }} + \dfrac{1}{{2021}} + \dfrac{{2020}}{{2021}}\\
    = {\left( {t\sqrt {2021}  – \dfrac{1}{{\sqrt {2021} }}} \right)^2} + \dfrac{{2020}}{{2021}}\\
   Do:{\left( {t\sqrt {2021}  – \dfrac{1}{{\sqrt {2021} }}} \right)^2} \ge 0\forall t\\
    \to {\left( {t\sqrt {2021}  – \dfrac{1}{{\sqrt {2021} }}} \right)^2} + \dfrac{{2020}}{{2021}} \ge \dfrac{{2020}}{{2021}}\\
    \to Min = \dfrac{{2020}}{{2021}}\\
    \Leftrightarrow t\sqrt {2021}  – \dfrac{1}{{\sqrt {2021} }} = 0\\
    \to t = \dfrac{1}{{2021}}\\
    \to \dfrac{1}{x} = \dfrac{1}{{2021}}\\
    \to x = 2021\\
   b)D = \dfrac{{{x^2} – 2x + 2015}}{{2015{x^2}}}\\
    = \dfrac{1}{{2015}} – \dfrac{2}{{2015x}} + \dfrac{1}{{{x^2}}}\\
   Đặt:\dfrac{1}{x} = t\\
    \to D = {t^2} – 2.t.\dfrac{1}{{2015}} + \dfrac{1}{{{{2015}^2}}} + \dfrac{1}{{2015}} – \dfrac{1}{{{{2015}^2}}}\\
    = {\left( {t – \dfrac{1}{{2015}}} \right)^2} + \dfrac{1}{{2015}} – \dfrac{1}{{{{2015}^2}}}\\
   Do:{\left( {t – \dfrac{1}{{2015}}} \right)^2} \ge 0\forall t\\
    \to {\left( {t – \dfrac{1}{{2015}}} \right)^2} + \dfrac{1}{{2015}} – \dfrac{1}{{{{2015}^2}}} \ge \dfrac{1}{{2015}} – \dfrac{1}{{{{2015}^2}}}\\
    \to Min = \dfrac{1}{{2015}} – \dfrac{1}{{{{2015}^2}}}\\
    \Leftrightarrow t = \dfrac{1}{{2015}}\\
    \to \dfrac{1}{x} = \dfrac{1}{{2015}}\\
    \to x = 2015
   \end{array}\)

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