Đáp án:
d) \(Max = \dfrac{1}{{12}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3;2;4} \right\}\\
A = \left[ {\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - x - 3}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}} \right].\left[ {\dfrac{{{x^2} + 5x + 14 - {x^2} - 4x - 16}}{{\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}}} \right]\\
= \dfrac{{{x^2} - 4 - x - 8}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}.\dfrac{{x - 2}}{{\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}}\\
= \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)\left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)}}\\
= \dfrac{1}{{{x^2} + 4x + 16}}\\
b)Thay:x = - 3\\
\to A = \dfrac{1}{{{{\left( { - 3} \right)}^2} + 4.\left( { - 3} \right) + 16}} = \dfrac{1}{{13}}\\
c)A = \dfrac{1}{{5x + 18}}\\
\to \dfrac{1}{{{x^2} + 4x + 16}} = \dfrac{1}{{5x + 18}}\left( {x \ne - \dfrac{{18}}{5}} \right)\\
\to 5x + 18 = {x^2} + 4x + 16\\
\to {x^2} - x - 2 = 0\\
\to \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = - 1\left( {TM} \right)
\end{array} \right.\\
d)A = \dfrac{1}{{{x^2} + 4x + 16}}\\
= \dfrac{1}{{{x^2} + 4x + 4 + 12}} = \dfrac{1}{{{{\left( {x + 2} \right)}^2} + 12}}\\
Do:{\left( {x + 2} \right)^2} \ge 0\forall x\\
\to {\left( {x + 2} \right)^2} + 12 \ge 12\\
\to \dfrac{1}{{{{\left( {x + 2} \right)}^2} + 12}} \le \dfrac{1}{{12}}\\
\to Max = \dfrac{1}{{12}}\\
\Leftrightarrow x + 2 = 0\\
\to x = - 2
\end{array}\)
