Đáp án:
\(\begin{array}{l}
a,\\
A = \dfrac{1}{4}\\
b,\\
B = \dfrac{{x - 4}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
c,\\
\left\{ \begin{array}{l}
0 \le x < 64\\
x \ne 4\\
x \ne 16
\end{array} \right.\\
d,\\
{E_{\min }} = - \dfrac{1}{2} \Leftrightarrow x = 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
x = 36 \Rightarrow A = \dfrac{{\sqrt {36} - 4}}{{\sqrt {36} + 2}} = \dfrac{{\sqrt {{6^2}} - 4}}{{\sqrt {{6^2}} + 2}} = \dfrac{{6 - 4}}{{6 + 2}} = \dfrac{2}{8} = \dfrac{1}{4}\\
b,\\
B = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 4}} + \dfrac{4}{{\sqrt x - 4}}} \right):\dfrac{{x + 16}}{{x - 4}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 4} \right) + 4.\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}:\dfrac{{x + 16}}{{x - 4}}\\
= \dfrac{{x - 4\sqrt x + 4\sqrt x + 16}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}:\dfrac{{x + 16}}{{x - 4}}\\
= \dfrac{{x + 16}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}.\dfrac{{x - 4}}{{x + 16}}\\
= \dfrac{{x - 4}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
c,\\
A.B = \dfrac{{\sqrt x - 4}}{{\sqrt x + 2}}.\dfrac{{x - 4}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{1}{{\sqrt x + 2}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x + 4}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 4}}\\
A.B < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{\sqrt x + 4}} < \dfrac{1}{2}\\
\Leftrightarrow \sqrt x - 2 < \dfrac{1}{2}\left( {\sqrt x + 4} \right)\\
\Leftrightarrow \sqrt x - 2 < \dfrac{1}{2}\sqrt x + 2\\
\Leftrightarrow \dfrac{1}{2}\sqrt x < 4\\
\Leftrightarrow \sqrt x < 8\\
\Leftrightarrow x < 64\\
\Rightarrow \left\{ \begin{array}{l}
0 \le x < 64\\
x \ne 4\\
x \ne 16
\end{array} \right.\\
d,\\
E = A.B = \dfrac{{\sqrt x - 2}}{{\sqrt x + 4}} = \dfrac{{\left( {\sqrt x + 4} \right) - 6}}{{\sqrt x + 4}} = 1 - \dfrac{6}{{\sqrt x + 4}}\\
\sqrt x + 4 \ge 4,\,\,\,\forall x \ge 0\\
\Rightarrow \dfrac{6}{{\sqrt x + 4}} \le \dfrac{6}{4} = \dfrac{3}{2},\,\,\,\,\forall x \ge 0\\
\Rightarrow E = 1 - \dfrac{6}{{\sqrt x + 4}} \ge 1 - \dfrac{3}{2} = - \dfrac{1}{2},\,\,\,\forall x \ge 0\\
\Rightarrow {E_{\min }} = - \dfrac{1}{2} \Leftrightarrow x = 0
\end{array}\)
