Đáp án:
h) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
b)A = \left[ {\dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]:\dfrac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}\\
= \dfrac{{2x - 2x - 4 - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{6}\\
= - \dfrac{6}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{6}\\
= - \dfrac{1}{{x - 2}}\\
c)\left| x \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = - \dfrac{1}{{\dfrac{1}{2} - 2}} = \dfrac{2}{3}\\
A = - \dfrac{1}{{ - \dfrac{1}{2} - 2}} = \dfrac{2}{5}
\end{array} \right.\\
d)A = 2\\
\to - \dfrac{1}{{x - 2}} = 2\\
\to x - 2 = - \dfrac{1}{2}\\
\to x = \dfrac{3}{2}\\
e)A > 0\\
\to - \dfrac{1}{{x - 2}} > 0\\
\to x - 2 < 0\\
\to x < 2;x \ne - 2\\
h)A \in Z\\
\to - \dfrac{1}{{x - 2}} \in Z\\
\to x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
