Đáp án:
$\begin{array}{l}
1)\dfrac{2}{{\sqrt 3 + 1}} - \dfrac{3}{{2\sqrt 3 + 3}} + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} \\
= \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} - \dfrac{{\sqrt 3 }}{{2 - \sqrt 3 }} + \sqrt 3 - 1\\
= \sqrt 3 - 1 - \dfrac{{\sqrt 3 \left( {2 + \sqrt 3 } \right)}}{{4 - 3}} + \sqrt 3 - 1\\
= 2\sqrt 3 - 2 - \sqrt 3 \left( {2 + \sqrt 3 } \right)\\
= 2\sqrt 3 - 2 - 2\sqrt 3 - 3\\
= - 5\\
2)\dfrac{{x - 4}}{{\sqrt x - 2}} - \dfrac{{x + 2\sqrt x }}{{\sqrt x + 2}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x - 2}} - \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 2 - \sqrt x \\
= 2\\
B2)Dkxd:2x - 1 \ge 0\\
\Rightarrow x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = 5\\
\Rightarrow 2x - 1 = 25\\
\Rightarrow 2x = 26\\
\Rightarrow x = 13\left( {tmdk} \right)\\
Vậy\,x = 13
\end{array}$
