Đáp án:
$\begin{array}{l}
5)0 \le x \le 4\\
\sqrt x + \sqrt {4 - x} = 6\\
\Rightarrow {\left( {\sqrt x + \sqrt {4 - x} } \right)^2} = 36\\
\Rightarrow x + 2\sqrt x .\sqrt {4 - x} + 4 - x = 36\\
\Rightarrow 4 + 2\sqrt {4x - {x^2}} = 36\\
\Rightarrow - 2\sqrt {4x - {x^2}} = - 32\\
\Rightarrow A = \sqrt {4 - 2\sqrt {4x - {x^2}} } = \sqrt {4 - \left( { - 32} \right)} = 6\\
6)0 \le x \le 9\\
\sqrt x + \sqrt {9 - x} = \sqrt {15} \\
\Rightarrow x + 2\sqrt {9x - {x^2}} + 9 - x = 15\\
\Rightarrow 2\sqrt {9x - {x^2}} = 6\\
\Rightarrow B = \sqrt {9 + 2\sqrt {9x - {x^2}} } = \sqrt {9 + 6} = \sqrt {15} \\
4)Dkxd:x > 0;x \ne 4\\
D = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 2}} + \dfrac{{\sqrt x }}{{\sqrt x + 2}}} \right).\dfrac{{x - 4}}{{2\sqrt x }}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{2\sqrt x }}\\
= \dfrac{{x + 2\sqrt x + x - 2\sqrt x }}{{2\sqrt x }}\\
= \dfrac{{2x}}{{2\sqrt x }}\\
= \sqrt x \\
5)Dkxd:x > 0;x \ne 1;x \ne 4\\
E = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }}} \right).\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\
= \dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x - 1 + x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x - 5}}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}.\left( {\sqrt x - 2} \right)}}
\end{array}$
