Đáp án:
$\begin{array}{l}
B2)a)B = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{x}{{x - 1}}} \right):\dfrac{{\sqrt x }}{{1 - \sqrt x }}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + x}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{1 - \sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - \sqrt x + x}}{{\sqrt x + 1}}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{2x - \sqrt x }}{{\sqrt x + 1}}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{1 - 2\sqrt x }}{{\sqrt x + 1}}\\
b)B = - 1\\
\Leftrightarrow \dfrac{{1 - 2\sqrt x }}{{\sqrt x + 1}} = - 1\\
\Leftrightarrow 2\sqrt x - 1 = \sqrt x + 1\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
B3)Dkxd:x > 0\\
a)A = \left( {\dfrac{9}{{x + 3\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x + 3}}} \right).\dfrac{{\sqrt x }}{2}\\
= \dfrac{{9 - \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x }}{2}\\
= \dfrac{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}{{\sqrt x + 3}}.\dfrac{1}{2}\\
= \dfrac{{3 - \sqrt x }}{2}\\
b)A > 0\\
\Leftrightarrow \dfrac{{3 - \sqrt x }}{2} > 0\\
\Leftrightarrow 3 - \sqrt x > 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
Vậy\,0 < x < 9\\
B4)\\
a)P = \sqrt {45} - \sqrt {16{{\left( {2 - \sqrt 5 } \right)}^2}} + \sqrt {6 + 2\sqrt 5 } \\
= 3\sqrt 5 - 4\left( {\sqrt 5 - 2} \right) + \sqrt 5 + 1\\
= 3\sqrt 5 - 4\sqrt 5 + 8 + \sqrt 5 + 1\\
= 9\\
b)Dkxd:x > 0\\
B = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 2}} - \dfrac{4}{{x + 2\sqrt x }}} \right).\dfrac{{\sqrt x }}{5}\\
= \dfrac{{\sqrt x .\sqrt x - 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x }}{5}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x + 2}}.\dfrac{1}{5}\\
= \dfrac{{\sqrt x - 2}}{5}\\
B < 1\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{5} < 1\\
\Leftrightarrow \dfrac{{\sqrt x - 2 - 5}}{5} < 0\\
\Leftrightarrow \sqrt x - 7 < 0\\
\Leftrightarrow \sqrt x < 7\\
\Leftrightarrow x < 49\\
Vậy\,0 < x < 49
\end{array}$
