Tìm GTLN / GTNN ạ)P=4x-2 √x +5 b) x- √x +1

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1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   P = 4x – 2\sqrt x  + 5\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
    = \left( {4x – 2\sqrt x  + \dfrac{1}{4}} \right) + \dfrac{{19}}{4}\\
    = \left( {{{\left( {2\sqrt x } \right)}^2} – 2.2\sqrt x .\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{{19}}{4}\\
    = {\left( {2\sqrt x  – \dfrac{1}{2}} \right)^2} + \dfrac{{19}}{4} \ge \dfrac{{19}}{4},\,\,\,\,\forall x \ge 0\\
    \Rightarrow {P_{\min }} = \dfrac{{19}}{4} \Leftrightarrow {\left( {2\sqrt x  – \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow \sqrt x  = \dfrac{1}{4} \Leftrightarrow x = \dfrac{1}{{16}}\\
   b,\\
   A = x – \sqrt x  + 1\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
    = \left( {x – \sqrt x  + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
    = \left( {x – 2.\sqrt x .\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4}\\
    = {\left( {\sqrt x  – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\,\forall x \ge 0\\
    \Rightarrow {A_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {\sqrt x  – \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{4}
   \end{array}\)

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