Question

Đã trả lời bởi chuyên gia

Answer 1

\(\begin{array}{l}
A = (\frac{{{x^2} - 3}}{{{x^2} - 9}} + \frac{1}{{x - 3}}):\frac{x}{{x + 3}}\\
dk:x \ne 3;x \ne - 3;x \ne 0\\
a)A = [\frac{{{x^2} - 3}}{{(x - 3)(x + 3)}} + \frac{{x + 3}}{{(x - 3)(x + 3)}}].\frac{{x + 3}}{x}\\
= \frac{{{x^2} - 3 + x + 3}}{{(x - 3)(x + 3)}}.\frac{{x + 3}}{x}\\
= \frac{{x(x + 1)}}{{x - 3}}.\frac{1}{x}\\
= \frac{{x + 1}}{{x - 3}}\\
b)|A| = 3\\
= > |\frac{{x + 1}}{{x - 3}}| = 3(x \ne 3;x \ne - 3;x \ne 0)\\
= > \left[ \begin{array}{l}
\frac{{x + 1}}{{x - 3}} = 3\\
\frac{{x + 1}}{{x - 3}} = - 3
\end{array} \right.\\
= > \left[ \begin{array}{l}
x + 1 = 3x - 9\\
x + 1 = - 3x + 9
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = 5\\
x = 2
\end{array} \right.(tm)\\
c)A = \frac{{x + 1}}{{x - 3}}(x \ne 3;x \ne - 3;x \ne 0)\\
= \frac{{x - 3 + 4}}{{x - 3}}\\
= 1 + \frac{4}{{x - 3}}\\
A \in Z \Leftrightarrow \frac{4}{{x - 3}} \in Z\\
= > x - 3 \in U(4) = \{ - 4; - 2; - 1;1;2;4\} \\
= > x \in \{ - 1;1;2;4;5;7\} (tm)\\
d)A > 0 = > \frac{{x + 1}}{{x - 3}} > 0(x \ne 3;x \ne - 3;x \ne 0)\\
= > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 > 0\\
x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + < 0\\
x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
= > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x > - 1\\
x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x < - 1\\
x < 3
\end{array} \right.}
\end{array}} \right.\\
= > \left[ \begin{array}{l}
x > 3\\
x < - 1
\end{array} \right.\\
= > \left[ \begin{array}{l}
x > 3\\
x < - 1;x \ne - 3
\end{array} \right.
\end{array}\)

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Answer 2

\(\begin{array}{l}
A = (\frac{{{x^2} - 3}}{{{x^2} - 9}} + \frac{1}{{x - 3}}):\frac{x}{{x + 3}}\\
dk:x \ne 3;x \ne - 3;x \ne 0\\
a)A = [\frac{{{x^2} - 3}}{{(x - 3)(x + 3)}} + \frac{{x + 3}}{{(x - 3)(x + 3)}}].\frac{{x + 3}}{x}\\
= \frac{{{x^2} - 3 + x + 3}}{{(x - 3)(x + 3)}}.\frac{{x + 3}}{x}\\
= \frac{{x(x + 1)}}{{x - 3}}.\frac{1}{x}\\
= \frac{{x + 1}}{{x - 3}}\\
b)|A| = 3\\
= > |\frac{{x + 1}}{{x - 3}}| = 3(x \ne 3;x \ne - 3;x \ne 0)\\
= > \left[ \begin{array}{l}
\frac{{x + 1}}{{x - 3}} = 3\\
\frac{{x + 1}}{{x - 3}} = - 3
\end{array} \right.\\
= > \left[ \begin{array}{l}
x + 1 = 3x - 9\\
x + 1 = - 3x + 9
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = 5\\
x = 2
\end{array} \right.(tm)\\
c)A = \frac{{x + 1}}{{x - 3}}(x \ne 3;x \ne - 3;x \ne 0)\\
= \frac{{x - 3 + 4}}{{x - 3}}\\
= 1 + \frac{4}{{x - 3}}\\
A \in Z \Leftrightarrow \frac{4}{{x - 3}} \in Z\\
= > x - 3 \in U(4) = \{ - 4; - 2; - 1;1;2;4\} \\
= > x \in \{ - 1;1;2;4;5;7\} (tm)\\
d)A > 0 = > \frac{{x + 1}}{{x - 3}} > 0(x \ne 3;x \ne - 3;x \ne 0)\\
= > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 > 0\\
x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + < 0\\
x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
= > \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x > - 1\\
x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x < - 1\\
x < 3
\end{array} \right.}
\end{array}} \right.\\
= > \left[ \begin{array}{l}
x > 3\\
x < - 1
\end{array} \right.\\
= > \left[ \begin{array}{l}
x > 3\\
x < - 1;x \ne - 3
\end{array} \right.
\end{array}\)

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