Đáp án:
\(MaxG = \dfrac{{121}}{{20}}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)B = {\left( {2x} \right)^2} + 2.2x.1 + 1\\
= {\left( {2x + 1} \right)^2}\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x \in R\\
\to B \ge 0\\
\to MinB = 0\\
\Leftrightarrow 2x + 1 = 0\\
\Leftrightarrow x = - \dfrac{1}{2}\\
3)C = - \left( {9{x^2} - 2.3x.5 + 25 - 21} \right)\\
= - {\left( {3x - 5} \right)^2} + 21\\
Do:{\left( {3x - 5} \right)^2} \ge 0\forall x\\
\to - {\left( {3x - 5} \right)^2} \le 0\\
\to - {\left( {3x - 5} \right)^2} + 21 \le 21\\
\to MaxC = 21\\
\Leftrightarrow 3x - 5 = 0\\
\to x = \dfrac{5}{3}\\
4)D = {y^2} - 2y.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {y - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {y - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {y - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to MinD = \dfrac{3}{4}\\
\Leftrightarrow y - \dfrac{1}{2} = 0\\
\Leftrightarrow y = \dfrac{1}{2}\\
5)E = - \left( {{x^2} - 2x + 1 + 2} \right)\\
= - {\left( {x - 1} \right)^2} - 2\\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 1} \right)^2} \le 0\\
\to - {\left( {x - 1} \right)^2} - 2 \le - 2\\
\to MaxE = - 2\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
6)F = {y^2} + 2y.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{25}}{4}\\
= {\left( {y + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4}\\
Do:{\left( {y + \dfrac{5}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {y + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} \ge \dfrac{{25}}{4}\\
\to MinF = \dfrac{{25}}{4}\\
\Leftrightarrow y + \dfrac{5}{2} = 0\\
\Leftrightarrow y = - \dfrac{5}{2}\\
7)G = - \left( {5{y^2} - 2.y\sqrt 5 .\dfrac{{11}}{{2\sqrt 5 }} + {{\left( {\dfrac{{11}}{{2\sqrt 5 }}} \right)}^2} - {{\left( {\dfrac{{11}}{{2\sqrt 5 }}} \right)}^2}} \right)\\
= - {\left( {y\sqrt 5 - \dfrac{{11}}{{2\sqrt 5 }}} \right)^2} + \dfrac{{121}}{{20}}\\
Do:{\left( {y\sqrt 5 - \dfrac{{11}}{{2\sqrt 5 }}} \right)^2} \ge 0\forall x\\
\to - {\left( {y\sqrt 5 - \dfrac{{11}}{{2\sqrt 5 }}} \right)^2} \le 0\\
\to - {\left( {y\sqrt 5 - \dfrac{{11}}{{2\sqrt 5 }}} \right)^2} + \dfrac{{121}}{{20}} \le \dfrac{{121}}{{20}}\\
\to MaxG = \dfrac{{121}}{{20}}\\
\Leftrightarrow y\sqrt 5 - \dfrac{{11}}{{2\sqrt 5 }} = 0\\
\Leftrightarrow y = \dfrac{{11}}{{10}}
\end{array}\)
