Đáp án:
$\begin{array}{l}
1)x = \sqrt {{{\left( {5 + \sqrt {13} } \right)}^2}} + \sqrt {{{\left( {\sqrt {13} - 4} \right)}^2}} = 5 + \sqrt {13} + 4 - \sqrt {13} = 9\\
\Rightarrow \sqrt x = \sqrt 9 = 3\\
\Rightarrow A = \frac{{x + \sqrt x + 1}}{{x + 1}} = \frac{{9 + 3 + 1}}{{9 + 1}} = \frac{{13}}{{10}}\\
2)Dkxd:x \ge 0;x \ne 1\\
B = \frac{1}{{\sqrt x - 1}} - \frac{{2\sqrt x }}{{x\sqrt x + \sqrt x - x - 1}}\\
= \frac{1}{{\sqrt x - 1}} - \frac{{2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{x + 1 - 2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x - 1}}{{x + 1}}
\end{array}$
