Rút gọn các biểu thức :
a) \(\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right)\left( {\sqrt a - \dfrac{1}{a}} \right)\);
b) \(\left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{x}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\);
c) \(\dfrac{{\sqrt x + 7x + 13}}{{x + 3\sqrt x - 10}} + \dfrac{{\sqrt x + 5}}{{2 - \sqrt x }} - \dfrac{{\sqrt x - 4}}{{\sqrt x + 5}}\).
d) \(\left( {\dfrac{{\left( {16 - \sqrt a } \right)\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a + 2}}} \right):\dfrac{1}{{a + 4\sqrt a + 4}}\).
+) Tìm điều kiện của x để biểu thức A xác định.
+) Quy đồng mẫu các phân thức sau đó biến đổi để rút gọn biểu thức.
\(a)\;\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right)\left( {\sqrt a - \dfrac{1}{a}} \right)\)
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Điều kiện: \(\left\{ \begin{array}{l}a \ge 0\\a \ne 0\\\sqrt a - 1 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a > 0\\a \ne 1\end{array} \right..\)
\(\begin{array}{l}\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right)\left( {\sqrt a - \dfrac{1}{a}} \right) = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2} - {{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}\left( {\dfrac{{a\sqrt a - 1}}{a}} \right)\\ = \dfrac{{a - 2\sqrt a + 1 - a - 2\sqrt a - 1}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{a}\\ = \dfrac{{ - 4\sqrt a }}{{\sqrt a + 1}}.\dfrac{{a + \sqrt a + 1}}{a}\\ = \dfrac{{ - 4\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}} = - \dfrac{{4a + 4\sqrt a + 4}}{{a + \sqrt a }} = - \left( {4 + \dfrac{4}{{a + \sqrt a }}} \right).\end{array}\)
\(b)\;\left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{x}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\)
Điều kiện: \(\left\{ \begin{array}{l}x \ge 0\\x \ne 0\\\sqrt x - 2 \ne 0\\\sqrt x - 1 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 0\\x \ne 4\\x \ne 1\end{array} \right..\)
\(\begin{array}{l}\;\;\;\left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{x}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\ = \dfrac{{\sqrt x - 1}}{x}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\ = \dfrac{{\sqrt x - 1}}{x}:\dfrac{{x - 1 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\ = \dfrac{{\sqrt x - 1}}{x}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{3}\\ = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}\left( {\sqrt x - 2} \right)}}{{3x}}.\end{array}\)
\(\begin{array}{l}c)\;\dfrac{{\sqrt x + 7x + 13}}{{x + 3\sqrt x - 10}} + \dfrac{{\sqrt x + 5}}{{2 - \sqrt x }} - \dfrac{{\sqrt x - 4}}{{\sqrt x + 5}}\\C = \;\dfrac{{\sqrt x + 7x + 13}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 5} \right)}} - \dfrac{{\sqrt x + 5}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 4}}{{\sqrt x + 5}}\end{array}\)
Điều kiện: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x - 2 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 4\end{array} \right..\)
\(\begin{array}{l}C = \;\dfrac{{\sqrt x + 7x + 13}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 5} \right)}} - \dfrac{{\sqrt x + 5}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 4}}{{\sqrt x + 5}}\\ = \dfrac{{\sqrt x + 7x + 13 - {{\left( {\sqrt x + 5} \right)}^2} - \left( {\sqrt x - 4} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 5} \right)}}\\ = \dfrac{{7x + \sqrt x + 13 - x - 10\sqrt x - 25 - x + 6\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 5} \right)}}\\ = \dfrac{{5x - 3\sqrt x - 20}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 5} \right)}}\end{array}\)
\(d)\;\left( {\dfrac{{\left( {16 - \sqrt a } \right)\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a + 2}}} \right):\dfrac{1}{{a + 4\sqrt a + 4}}\)
Điều kiện: \(\left\{ \begin{array}{l}a \ge 0\\2 - \sqrt a \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge 0\\a \ne 4\end{array} \right..\)
\(\begin{array}{l}\left( {\dfrac{{\left( {16 - \sqrt a } \right)\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a + 2}}} \right):\dfrac{1}{{a + 4\sqrt a + 4}}\\ = \dfrac{{16\sqrt a - a - \left( {2\sqrt a + 3} \right)\left( {\sqrt a + 2} \right) - \left( {2 - 3\sqrt a } \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}:\dfrac{1}{{{{\left( {\sqrt a + 2} \right)}^2}}}\\ = \dfrac{{16\sqrt a - a - 2a - 7\sqrt a - 6 + 4 - 8\sqrt a + 3a}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}.{\left( {\sqrt a + 2} \right)^2}\\ = \dfrac{{\sqrt a }}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{\left( {\sqrt a + 2} \right)^2} = \dfrac{{\sqrt a \left( {\sqrt a + 2} \right)}}{{\sqrt a - 2}}.\end{array}\)