Bài 5 trang 39 Tài liệu dạy – học Toán 9 tập 1

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) \(\left( {\dfrac{{\left( {16 - \sqrt a } \right)\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a  + 2}}} \right):\dfrac{1}{{a + 4\sqrt a  + 4}}\).

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: \(\left\{ \begin{array}{l}a \ge 0\\2 - \sqrt a  \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge 0\\a \ne 4\end{array} \right..\)

\(\begin{array}{l}\left( {\dfrac{{\left( {16 - \sqrt a } \right)\sqrt a }}{{a - 4}} + \dfrac{{3 + 2\sqrt a }}{{2 - \sqrt a }} - \dfrac{{2 - 3\sqrt a }}{{\sqrt a  + 2}}} \right):\dfrac{1}{{a + 4\sqrt a  + 4}}\\ = \dfrac{{16\sqrt a  - a - \left( {2\sqrt a  + 3} \right)\left( {\sqrt a  + 2} \right) - \left( {2 - 3\sqrt a } \right)\left( {\sqrt a  - 2} \right)}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  + 2} \right)}}:\dfrac{1}{{{{\left( {\sqrt a  + 2} \right)}^2}}}\\ = \dfrac{{16\sqrt a  - a - 2a - 7\sqrt a  - 6 + 4 - 8\sqrt a  + 3a}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  + 2} \right)}}.{\left( {\sqrt a  + 2} \right)^2}\\ = \dfrac{{\sqrt a }}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  + 2} \right)}}{\left( {\sqrt a  + 2} \right)^2} \\= \dfrac{{\sqrt a \left( {\sqrt a  + 2} \right)}}{{\sqrt a  - 2}}.\end{array}\)