Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.DK:x \ne \pm 2\\
A = \frac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\frac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}\\
= \frac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{x + 2}}{6}\\
= \frac{{ - 1}}{{x - 2}}\\
b.\left| x \right| = \frac{1}{2}\\
\to \left[ \begin{array}{l}
x = \frac{1}{2}\\
x = - \frac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \frac{2}{3}\\
A = \frac{2}{5}
\end{array} \right.\\
c.A.\left| {x - 3} \right| < 0\\
\to A < 0\left( {do:\left| {x - 3} \right| \ge 0\forall x \in R} \right)\\
\to \frac{{ - 1}}{{x - 2}} < 0\\
\to x - 2 > 0\\
\to x > 2\\
d. \Leftrightarrow \left( {x - 2} \right) \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\left( {TM} \right)
\end{array}\)
