Đáp án:
a) Theo Pytago ta có:
$\begin{array}{l}
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow B{C^2} = {5^2} + {12^2} = 169\\
\Rightarrow BC = 13\left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.AH.BC\\
\Rightarrow AH = \dfrac{{5.12}}{{13}} = \dfrac{{60}}{{13}}\left( {cm} \right)\\
b)\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{12}}{{13}}\\
\Rightarrow \widehat B = {67^0}\\
\Rightarrow \widehat C = {90^0} - {67^0} = {23^0}\\
Vậy\,\widehat C = {23^0}\\
B2)\\
a)\dfrac{3}{{2 - \sqrt 5 }} - \dfrac{3}{{2 + \sqrt 5 }}\\
= \dfrac{{3\left( {2 + \sqrt 5 } \right)}}{{4 - 5}} - \dfrac{{3\left( {2 - \sqrt 5 } \right)}}{{4 - 5}}\\
= - 6 - 3\sqrt 5 + 6 - 3\sqrt 5 \\
= - 6\sqrt 5 \\
b)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{1 - \sqrt x }}{{1 + \sqrt x }} - \dfrac{{1 + \sqrt x }}{{1 - \sqrt x }}\\
= \dfrac{{{{\left( {1 - \sqrt x } \right)}^2} - {{\left( {1 + \sqrt x } \right)}^2}}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{1 - 2\sqrt x + x - \left( {1 + 2\sqrt x + x} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{ - 4\sqrt x }}{{1 - x}}\\
= \dfrac{{4\sqrt x }}{{x - 1}}
\end{array}$
