Đáp án:
$\begin{array}{l}
9)Dkxd:x \ge 0;x \ne 4\\
P = \left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{\sqrt x + 2}} + \dfrac{4}{{x - 4}}} \right):\dfrac{{\sqrt x + 1}}{{x - 4}}\\
= \dfrac{{\sqrt x + 2 - \left( {\sqrt x - 2} \right) + 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{x - 4}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 2 - \sqrt x + 2 + 4}}{{x - 4}}.\dfrac{{x - 4}}{{\sqrt x + 1}}\\
= \dfrac{8}{{\sqrt x + 1}}\\
10)a)A{H^2} = BH.CH = 2.4 = 8\\
\Rightarrow AH = \sqrt 8 = 2\sqrt 2 \left( {cm} \right)\\
Trong:\Delta ABH \bot H\\
\Rightarrow \tan \widehat B = \dfrac{{AH}}{{BH}} = \dfrac{{2\sqrt 2 }}{2} = \sqrt 2 \\
\Rightarrow \widehat B = {51^0}\\
\Rightarrow \widehat C = {90^0} - \widehat B = {90^0} - {51^0} = {39^0}\\
\text{Vậy}\,\widehat B = {51^0};\widehat C = {39^0}\\
b)AH = 2\sqrt 2 \left( {cm} \right)\\
Theo\,Pytago:\\
A{B^2} = A{H^2} + B{H^2}\\
= {2^2} + {\left( {2\sqrt 2 } \right)^2} = 12\\
\Rightarrow AB = 2\sqrt 3 \left( {cm} \right)\\
\text{Vậy}\,AB = 2\sqrt 3 cm;AH = 2\sqrt 2 cm
\end{array}$
