Đáp án:
$\begin{array}{l}
B1)Dkxd:\left\{ \begin{array}{l}
{x^2} - 5 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ge 5\\
x \ge 1
\end{array} \right. \Leftrightarrow x \ge \sqrt 5 \\
\sqrt {{x^2} - 5} = x - 1\\
\Leftrightarrow {x^2} - 5 = {x^2} - 2x + 1\\
\Leftrightarrow 2x = 6\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
Vậy\,x = 3\\
B2)\\
a)\dfrac{{x\sqrt x - y\sqrt y }}{{\sqrt x - \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x } \right)}^3} - {{\left( {\sqrt y } \right)}^3}}}{{\sqrt x - \sqrt y }}\\
= \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\sqrt x - \sqrt y }}\\
= x + \sqrt {xy} + y\\
b)\dfrac{{x\sqrt x - 1}}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x } \right)}^3} - 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= x + \sqrt x + 1
\end{array}$
