Đáp án:
c) \(\left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a)DK:x \ne \left\{ { - 3; - 2;2} \right\}\\
D = \dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - x - 3}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}\\
= \dfrac{{{x^2} - 4 - x - 8}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}\\
= \dfrac{{{x^2} - x - 12}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}\\
= \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}.\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}\\
= \dfrac{{x + 2}}{{x - 1}}\\
b)D = \dfrac{3}{2}\\
\to \dfrac{{x + 2}}{{x - 1}} = \dfrac{3}{2}\\
\to 2x + 4 = 3x - 3\\
\to x = 7\\
c)D = \dfrac{{x + 2}}{{x - 1}} = \dfrac{{x - 1 + 3}}{{x - 1}} = 1 + \dfrac{3}{{x - 1}}\\
D \in Z \to \dfrac{3}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - 2\left( l \right)\\
x = 2\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)
