Đáp án:
d) \(Min = - \dfrac{5}{7}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{2}{{{x^2} - 4x + 4 + 1}} = \dfrac{2}{{{{\left( {x - 2} \right)}^2} + 1}}\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x - 2} \right)^2} + 1 \ge 1\\
\to \dfrac{2}{{{{\left( {x - 2} \right)}^2} + 1}} \le 2\\
\to Max = 2\\
\Leftrightarrow x = 2\\
B = \dfrac{3}{{4{x^2} + 2.2x.\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{31}}{{16}}}}\\
= \dfrac{3}{{{{\left( {2x + \dfrac{1}{4}} \right)}^2} + \dfrac{{31}}{{16}}}}\\
Do:{\left( {2x + \dfrac{1}{4}} \right)^2} \ge 0\\
\to {\left( {2x + \dfrac{1}{4}} \right)^2} + \dfrac{{31}}{{16}} \ge \dfrac{{31}}{{16}}\\
\to \dfrac{3}{{{{\left( {2x + \dfrac{1}{4}} \right)}^2} + \dfrac{{31}}{{16}}}} \le 3:\dfrac{{31}}{{16}}\\
\to \dfrac{3}{{{{\left( {2x + \dfrac{1}{4}} \right)}^2} + \dfrac{{31}}{{16}}}} \le \dfrac{{48}}{{31}}\\
\to Max = \dfrac{{48}}{{31}}\\
\Leftrightarrow 2x + \dfrac{1}{4} = 0\\
\Leftrightarrow x = - \dfrac{1}{8}\\
C = \dfrac{5}{{3{x^2} - 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}} + \dfrac{{71}}{{12}}}}\\
= \dfrac{5}{{{{\left( {x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }}} \right)}^2} + \dfrac{{71}}{{12}}}}\\
Do:{\left( {x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }}} \right)^2} \ge 0\\
\to {\left( {x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{71}}{{12}} \ge \dfrac{{71}}{{12}}\\
\to \dfrac{5}{{{{\left( {x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }}} \right)}^2} + \dfrac{{71}}{{12}}}} \le 5:\dfrac{{71}}{{12}}\\
\to \dfrac{5}{{{{\left( {x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }}} \right)}^2} + \dfrac{{71}}{{12}}}} \le \dfrac{{60}}{{71}}\\
\to Max = \dfrac{{60}}{{71}}\\
\Leftrightarrow x\sqrt 3 - \dfrac{1}{{2\sqrt 3 }} = 0\\
\Leftrightarrow x = \dfrac{1}{6}\\
D = \dfrac{5}{{ - \left( {9{x^2} - 6x + 8} \right)}} = - \dfrac{5}{{9{x^2} - 2.3x.1 + 1 + 7}}\\
= - \dfrac{5}{{{{\left( {3x - 1} \right)}^2} + 7}}\\
Do:{\left( {3x - 1} \right)^2} \ge 0\forall x\\
\to {\left( {3x - 1} \right)^2} + 7 \ge 7\\
\to \dfrac{5}{{{{\left( {3x - 1} \right)}^2} + 7}} \le \dfrac{5}{7}\\
\to - \dfrac{5}{{{{\left( {3x - 1} \right)}^2} + 7}} \ge - \dfrac{5}{7}\\
\to Min = - \dfrac{5}{7}\\
\Leftrightarrow 3x - 1 = 0\\
\Leftrightarrow x = \dfrac{1}{3}\\
E = \dfrac{4}{{ - \left( {{x^2} - 5x + 6} \right)}} = - \dfrac{4}{{{x^2} - 2x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{1}{4}}}\\
= - \dfrac{4}{{{{\left( {x - \dfrac{5}{2}} \right)}^2} - \dfrac{1}{4}}}\\
Do:{\left( {x - \dfrac{5}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x - \dfrac{5}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\to \dfrac{4}{{{{\left( {x - \dfrac{5}{2}} \right)}^2} - \dfrac{1}{4}}} \le - 16\\
\to - \dfrac{4}{{{{\left( {x - \dfrac{5}{2}} \right)}^2} - \dfrac{1}{4}}} \ge 16\\
\to Min = 16\\
\Leftrightarrow x = \dfrac{5}{2}
\end{array}\)
