Đề bài
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] \[\left[ {\dfrac{{\left[ {16 - \sqrt a } \right]\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a + 2}}} \right]:\dfrac{1}{{a + 4\sqrt a + 4}}\].
Phương pháp giải - Xem chi tiết
+] 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.
Lời giải chi tiết
\[a]\;\left[ {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right]\left[ {\sqrt a - \dfrac{1}{a}} \right]\]
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: \[\left\{ \begin{array}{l}a \ge 0\\2 - \sqrt a \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge 0\\a \ne 4\end{array} \right..\]
\[\begin{array}{l}\left[ {\dfrac{{\left[ {16 - \sqrt a } \right]\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a + 2}}} \right]:\dfrac{1}{{a + 4\sqrt a + 4}}\\ = \dfrac{{16\sqrt a - a - \left[ {2\sqrt a + 3} \right]\left[ {\sqrt a + 2} \right] - \left[ {2 - 3\sqrt a } \right]\left[ {\sqrt a - 2} \right]}}{{\left[ {\sqrt a - 2} \right]\left[ {\sqrt a + 2} \right]}}:\dfrac{1}{{{{\left[ {\sqrt a + 2} \right]}^2}}}\\ = \dfrac{{16\sqrt a - a - 2a - 7\sqrt a - 6 + 4 - 8\sqrt a + 3a}}{{\left[ {\sqrt a - 2} \right]\left[ {\sqrt a + 2} \right]}}.{\left[ {\sqrt a + 2} \right]^2}\\ = \dfrac{{\sqrt a }}{{\left[ {\sqrt a - 2} \right]\left[ {\sqrt a + 2} \right]}}{\left[ {\sqrt a + 2} \right]^2} \\= \dfrac{{\sqrt a \left[ {\sqrt a + 2} \right]}}{{\sqrt a - 2}}.\end{array}\]