D= 5X^2 + 3X + 18 E= ( x+ 1) ^2 + ( x+3 ) ^2 + 2( x-5) F + ( x+ 4)^ + ( 2x + 1) ^2 TÌMN GTNN

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1. Đáp án:

   $\begin{array}{l}
   D = 5{x^2} + 3x + 18\\
    = 5\left( {{x^2} + \dfrac{3}{5}x} \right) + 18\\
    = 5.\left( {{x^2} + 2.x.\dfrac{3}{{10}} + \dfrac{9}{{100}} – \dfrac{9}{{100}}} \right) + 18\\
    = 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} – 5.\dfrac{9}{{100}} + 18\\
    = 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} + \dfrac{{351}}{{20}}\\
   Do:5.{\left( {x + \dfrac{3}{{10}}} \right)^2} \ge 0\\
    \Rightarrow 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} + \dfrac{{351}}{{20}} \ge \dfrac{{351}}{{20}}\\
    \Rightarrow GTNN:D = \dfrac{{351}}{{20}} \Leftrightarrow x =  – \dfrac{3}{{10}}\\
   E = {\left( {x + 1} \right)^2} + {\left( {x + 3} \right)^2} + 2\left( {x – 5} \right)\\
    = 2{x^2} + 10x\\
    = 2\left( {{x^2} + 5x} \right)\\
    = 2\left( {{x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4}} \right) – 2.\dfrac{{25}}{4}\\
    = 2.{\left( {x + \dfrac{5}{2}} \right)^2} – \dfrac{{25}}{2} \ge  – \dfrac{{25}}{2}\\
    \Rightarrow GTNN:E =  – \dfrac{{25}}{2} \Leftrightarrow x =  – \dfrac{5}{2}\\
   F = {\left( {x + 4} \right)^2} + {\left( {2x + 1} \right)^2}\\
    = {x^2} + 8x + 16 + 4{x^2} + 4x + 1\\
    = 5{x^2} + 12x + 17\\
    = 5.\left( {{x^2} + \dfrac{{12}}{5}x} \right) + 17\\
    = 5.\left( {{x^2} + 2.x.\dfrac{6}{5} + \dfrac{{36}}{{25}}} \right) – 5.\dfrac{{36}}{{25}} + 17\\
    = 5.{\left( {x + \dfrac{6}{5}} \right)^2} + \dfrac{{49}}{5} \ge \dfrac{{49}}{5}\\
    \Rightarrow GTNN:F = \dfrac{{49}}{5} \Leftrightarrow x =  – \dfrac{6}{5}
   \end{array}$

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