Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {\dfrac{1}{{1 - \sqrt 3 }} - \dfrac{1}{{1 + \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 3 }}\\
= \dfrac{{\left( {1 + \sqrt 3 } \right) - \left( {1 - \sqrt 3 } \right)}}{{\left( {1 - \sqrt 3 } \right)\left( {1 + \sqrt 3 } \right)}}.\sqrt 3 \\
= \dfrac{{2\sqrt 3 }}{{1 - 3}}.\sqrt 3 \\
= \left( { - \sqrt 3 } \right).\sqrt 3 \\
= - 3\\
b,\\
\dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{{2\sqrt x - 1}}{{x - \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right)\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\sqrt x }^2} - \left( {2\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
2,\\
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x - 2 \ge 0\\
\sqrt {x - 2} \ne \sqrt 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \ne 5
\end{array} \right.\\
b,\\
P = \dfrac{{x - 5}}{{\sqrt {x - 2} - \sqrt 3 }} = \dfrac{{\left( {x - 5} \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}{{\left( {\sqrt {x - 2} - \sqrt 3 } \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}\\
= \dfrac{{\left( {x - 5} \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}{{\left( {x - 2} \right) - 3}} = \dfrac{{\left( {x - 5} \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}{{x - 5}}\\
= \sqrt {x - 2} + \sqrt 3 \\
c,\\
P = \dfrac{6}{5} \Leftrightarrow \sqrt {x - 2} + \sqrt 3 = \dfrac{6}{5}\\
\Leftrightarrow \sqrt {x - 2} = \dfrac{6}{5} - \sqrt 3 \\
\Leftrightarrow x - 2 = {\left( {\dfrac{6}{5} - \sqrt 3 } \right)^2}\\
\Leftrightarrow x - 2 = \dfrac{{111}}{{25}} - \dfrac{{12\sqrt 3 }}{5}\\
\Leftrightarrow x = \dfrac{{161}}{{25}} - \dfrac{{12\sqrt 3 }}{5}\\
\Leftrightarrow x = \dfrac{{161 - 60\sqrt 3 }}{{25}}\\
\end{array}\)
