Đáp án:
a) \(x \ne \pm 3;\,\,x \ne \pm 2.\)
b) \(D = \frac{3}{{x + 2}}.\)
c) \(x \in \left\{ { - 5;\,\, - 1;\,\,1} \right\}.\)
Giải thích các bước giải:
\(D = \left( {\frac{{{x^2} - 3x}}{{{x^2} - 9}} - 1} \right):\left[ {\frac{{9 - {x^2}}}{{\left( {x + 3} \right)\left( {x - 2} \right)}} + \frac{{x - 3}}{{x - 2}} - \frac{{x + 2}}{{x + 3}}} \right]\)
a) Điều kiện xác định:
\(\left\{ \begin{array}{l}{x^2} - 9 \ne 0\\\left( {x + 3} \right)\left( {x - 2} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {x - 3} \right)\left( {x + 3} \right) \ne 0\\x + 3 \ne 0\\x - 2 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 3\\x \ne - 3\\x \ne 2\end{array} \right..\)
b) Rút gọn:
\(\begin{array}{l}D = \left( {\frac{{{x^2} - 3x}}{{{x^2} - 9}} - 1} \right):\left[ {\frac{{9 - {x^2}}}{{\left( {x + 3} \right)\left( {x - 2} \right)}} + \frac{{x - 3}}{{x - 2}} - \frac{{x + 2}}{{x + 3}}} \right]\\ = \left[ {\frac{{x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - 1} \right]:\frac{{9 - {x^2} + \left( {x - 3} \right)\left( {x + 3} \right) - \left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\ = \left( {\frac{x}{{x + 3}} - 1} \right):\frac{{9 - {x^2} + {x^2} - 9 - {x^2} + 4}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\ = \frac{{x - x - 3}}{{x + 3}}:\frac{{ - \left( {{x^2} - 4} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\ = \frac{{ - 3}}{{x + 3}}.\frac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{ - \left( {x + 2} \right)\left( {x - 2} \right)}}\,\,\,\,\,\,\,\left( {x \ne - 2} \right)\\ = \frac{3}{{x + 2}}.\end{array}\)
c) ĐKXĐ: \(x \ne \pm 3;\,\,x \ne \pm 2.\)
Ta có: \(D = \frac{3}{{x + 2}}\)
\(\begin{array}{l} \Rightarrow D \in \mathbb{Z} \Leftrightarrow \frac{3}{{x + 2}} \in \mathbb{Z} \Rightarrow x + 2 \in U\left( 3 \right)\\ \Rightarrow x + 2 \in \left\{ { \pm 1;\,\, \pm 3} \right\}\\ \Rightarrow \left[ \begin{array}{l}x + 2 = - 3\\x + 2 = - 1\\x + 2 = 1\\x + 2 = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 5\,\,\,\left( {tm} \right)\\x = - 3\,\,\,\left( {ktm} \right)\\x = - 1\,\,\,\left( {tm} \right)\\x = 1\,\,\,\left( {tm} \right)\end{array} \right..\end{array}\)
Vậy \(x \in \left\{ { - 5;\,\, - 1;\,\,1} \right\}.\)
