D= ( x^3-3x / x^2 -9 -1 ) : [ 9-x^2/ ( x+ 3) (x-2) + x-3 /x-2 -x=2 /x+3]
a) tìm đk xác định
b) RG
c) tìm x nguyên để D nguyên
chi tiết dùm mk nha các bạn ơi
thankyou
D= ( x^3-3x / x^2 -9 -1 ) : [ 9-x^2/ ( x+ 3) (x-2) + x-3 /x-2 -x=2 /x+3]
a) tìm đk xác định
b) RG
c) tìm x nguyên để D nguyên
chi tiết dùm mk nha các bạn ơi
thankyou
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 \(x \in \left\{ { – 5;\,\, – 1;\,\,1} \right\}.\)