D = ( x^3 -3x/ x^2 -9 -1 ) : (9-x^2 /(x+ 3 ) ( x-2) + x-3/x-2 – x)
a) tìm đk xác định
b) RG
c) tìm x nguyên để D nguyên
D = ( x^3 -3x/ x^2 -9 -1 ) : (9-x^2 /(x+ 3 ) ( x-2) + x-3/x-2 – x)
a) tìm đk xác định
b) RG
c) tìm x nguyên để D nguyên
Đáp án:
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
D = \left( {\frac{{{x^2} – 3x}}{{{x^2} – 9}} – 1} \right):\left( {\frac{{9 – {x^2}}}{{\left( {x + 3} \right)\left( {x – 2} \right)}} + \frac{{x – 3}}{{x – 2}} – 1} \right)\\
a)Đkxđ:\left\{ \begin{array}{l}
{x^2} – 9 \ne 0\\
x + 3 \ne 0\\
x – 2 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 3\\
x \ne – 3\\
x \ne 2
\end{array} \right.\\
b)D = \left( {\frac{{{x^2} – 3x}}{{{x^2} – 9}} – 1} \right):\left( {\frac{{9 – {x^2}}}{{\left( {x + 3} \right)\left( {x – 2} \right)}} + \frac{{x – 3}}{{x – 2}} – 1} \right)\\
= \left( {\frac{{x\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} – 1} \right):\left( {\frac{{ – \left( {x – 3} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x – 2} \right)}} + \frac{{x – 3}}{{x – 2}} – 1} \right)\\
= \left( {\frac{x}{{x + 3}} – 1} \right):\left( {\frac{{3 – x}}{{x – 2}} + \frac{{x – 3}}{{x – 2}} – 1} \right)\\
= \frac{{x – x – 3}}{{x + 3}}:\frac{{3 – x + x – 3 – \left( {x – 2} \right)}}{{x – 2}}\\
= \frac{{ – 3}}{{x + 3}}:\left( { – 1} \right)\\
= \frac{3}{{\left( {x + 3} \right)}}\\
c)x \ne 2;x \ne 3;x \ne – 3\\
D = \frac{3}{{x + 3}}\\
D \in Z \Rightarrow \frac{3}{{x + 3}} \in Z\\
\Rightarrow \left( {x + 3} \right) \in U\left( 3 \right) = {\rm{\{ }} – 3; – 1;1;3\} \\
\Rightarrow x \in {\rm{\{ }} – 6; – 4; – 2;0\} \left( {tmdk} \right)
\end{array}$