Đáp án:
$\begin{array}{l}
1.4\\
a)P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
b)x = 4\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow P = \dfrac{{4 + 2 + 1}}{2} = \dfrac{7}{2}\\
c)P = \dfrac{{13}}{3}\\
\Leftrightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \dfrac{{13}}{3}\\
\Leftrightarrow 3x + 3\sqrt x + 3 = 13\sqrt x \\
\Leftrightarrow 3x - 10\sqrt x + 3 = 0\\
\Leftrightarrow \left( {3\sqrt x - 1} \right)\left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{1}{3}\\
\sqrt x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{9}\left( {tm} \right)\\
x = 9\left( {tm} \right)
\end{array} \right.\\
Vay\,x = \dfrac{1}{9};x = 9\\
1.5a)\dfrac{5}{{\sqrt 5 }} = \sqrt 5 \\
\dfrac{5}{{2 + \sqrt 3 }} = \dfrac{{5\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = 10 - 5\sqrt 3 \\
b)A = \dfrac{{\sqrt {ab} - 2\sqrt {{b^2}} }}{b} - \sqrt {\dfrac{a}{b}} \\
= \dfrac{{\sqrt {ab} - 2b}}{b} - \dfrac{{\sqrt {ab} }}{b}\\
= \dfrac{{\sqrt {ab} - 2b - \sqrt {ab} }}{b}\\
= \dfrac{{ - 2b}}{b}\\
= - 2
\end{array}$
