Đáp án:
a) \(\dfrac{{x - 2}}{{x - 4}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 4\\
D = \left[ {\dfrac{{\left( {x - 1} \right)\left( {x + 4} \right) + x\left( {x - 4} \right) - {x^2} + x}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}} \right]:\left( {\dfrac{{x + 4 - 2}}{{x + 4}}} \right)\\
= \dfrac{{{x^2} + 3x - 4 + {x^2} - 4x - {x^2} + x}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}.\dfrac{{x + 4}}{{x + 2}}\\
= \dfrac{{{x^2} - 4}}{{x - 4}}.\dfrac{1}{{x + 2}}\\
= \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x - 4} \right)\left( {x + 2} \right)}} = \dfrac{{x - 2}}{{x - 4}}\\
b)\left| {x - 3} \right| = 5\\
\to \left[ \begin{array}{l}
x - 3 = 5\\
x - 3 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
D = \dfrac{{8 - 2}}{{8 - 4}} = \dfrac{3}{2}\\
D = \dfrac{{ - 2 - 2}}{{ - 2 - 4}} = \dfrac{2}{3}
\end{array} \right.\\
c)D \ge 1\\
\to \dfrac{{x - 2}}{{x - 4}} \ge 1\\
\to \dfrac{{x - 2 - x + 4}}{{x - 4}} \ge 0\\
\to \dfrac{2}{{x - 4}} \ge 0\\
\to x - 4 > 0\\
\to x > 4\\
d)D = \dfrac{{x - 2}}{{x - 4}} = \dfrac{{x - 4 + 2}}{{x - 4}} = 1 + \dfrac{2}{{x - 4}}\\
D \in Z\\
\Leftrightarrow \dfrac{2}{{x - 4}} \in Z\\
\Leftrightarrow x - 4 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 4 = 2\\
x - 4 = - 2\\
x - 4 = 1\\
x - 4 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = 2\\
x = 5\\
x = 3
\end{array} \right.
\end{array}\)
