Đáp án:
\(\begin{array}{l}
2,\,\,\,\,\dfrac{{ - 2}}{{1 + \sqrt a }}\\
3,\,\,\,\,\dfrac{{2.\left( {a - 1} \right)}}{{a + 1}}\\
4,\,\,\,\, - 4\\
5,\,\,\,\,\dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
6,\,\,\,\,\dfrac{{1 - a}}{{\sqrt a }}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
\left( {\dfrac{1}{{1 - \sqrt a }} - \dfrac{1}{{1 + \sqrt a }}} \right).\left( {1 - \dfrac{1}{{\sqrt a }}} \right)\\
= \left( {\dfrac{{1.\left( {1 + \sqrt a } \right) - 1.\left( {1 - \sqrt a } \right)}}{{\left( {1 - \sqrt a } \right).\left( {1 + \sqrt a } \right)}}} \right).\dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
= \dfrac{{\left( {1 + \sqrt a } \right) - \left( {1 - \sqrt a } \right)}}{{\left( {1 - \sqrt a } \right).\left( {1 + \sqrt a } \right)}}.\dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
= \dfrac{{2\sqrt a }}{{\left( {1 - \sqrt a } \right).\left( {1 + \sqrt a } \right)}}.\dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
= \dfrac{{ - 2}}{{1 + \sqrt a }}\\
3,\\
\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} + \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right){\left( {1 - \dfrac{2}{{a + 1}}} \right)^2}\\
= \dfrac{{{{\left( {\sqrt a - 1} \right)}^2} + {{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.{\left( {\dfrac{{\left( {a + 1} \right) - 2}}{{a + 1}}} \right)^2}\\
= \dfrac{{\left( {a - 2\sqrt a + 1} \right) + \left( {a + 2\sqrt a + 1} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.{\left( {\dfrac{{a - 1}}{{a + 1}}} \right)^2}\\
= \dfrac{{2a + 2}}{{{{\sqrt a }^2} - {1^2}}}.{\left( {\dfrac{{a - 1}}{{a + 1}}} \right)^2}\\
= \dfrac{{2.\left( {a + 1} \right)}}{{a - 1}}.\dfrac{{{{\left( {a - 1} \right)}^2}}}{{{{\left( {a + 1} \right)}^2}}}\\
= \dfrac{{2.\left( {a - 1} \right)}}{{a + 1}}\\
4,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x + 1 \ne 0\\
\sqrt x - 1 \ne 0\\
\sqrt x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2} - {{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x }}\\
= \dfrac{{\left( {x - 2\sqrt x + 1} \right) - \left( {x + 2\sqrt x + 1} \right)}}{{{{\sqrt x }^2} - {1^2}}}.\dfrac{{x - 1}}{{\sqrt x }}\\
= \dfrac{{ - 4\sqrt x }}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x }}\\
= - 4\\
5,\\
\left( {\dfrac{1}{{a - \sqrt a }} + \dfrac{1}{{\sqrt a - 1}}} \right):\dfrac{{\sqrt a + 1}}{{a - 2\sqrt a + 1}}\\
= \left( {\dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}} + \dfrac{1}{{\sqrt a - 1}}} \right):\dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt a }}{{\sqrt a .\left( {\sqrt a - 1} \right)}}:\dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a .\left( {\sqrt a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
6,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
2\sqrt a \ne 0\\
\sqrt a + 1 \ne 0\\
\sqrt a - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 0\\
a \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\\
{\left( {\dfrac{{\sqrt a }}{2} - \dfrac{1}{{2\sqrt a }}} \right)^2}.\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right)\\
= {\left( {\dfrac{{\sqrt a .\sqrt a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2} - {{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right).\left( {\sqrt a - 1} \right)}}\\
= {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{\left( {a - 2\sqrt a + 1} \right) - \left( {a + 2\sqrt a + 1} \right)}}{{{{\sqrt a }^2} - {1^2}}}\\
= \dfrac{{{{\left( {a - 1} \right)}^2}}}{{4a}}.\dfrac{{ - 4\sqrt a }}{{a - 1}}\\
= \dfrac{{ - \left( {a - 1} \right)}}{{\sqrt a }}\\
= \dfrac{{1 - a}}{{\sqrt a }}
\end{array}\)