Đáp án:
B1:
a) \(\dfrac{{x - 4}}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne \left\{ { - 3;2} \right\}\\
M = \dfrac{{x + 2}}{{x + 3}} - \dfrac{5}{{{x^2} + x - 6}} + \dfrac{1}{{2 - x}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - x - 3}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 4 - x - 8}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - x - 12}}{{\left( {x - 2} \right)\left( {x + 3} \right)}} = \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x - 4}}{{x - 2}}\\
b)M = \dfrac{{x - 2 - 2}}{{x - 2}} = 1 - \dfrac{2}{{x - 2}}\\
M \in Z \to \dfrac{2}{{x - 2}} \in Z\\
\to x - 2 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0\\
x = 3\\
x = 1
\end{array} \right.\\
B2:\\
\left( {1 - 2xy} \right).\dfrac{{x + 2}}{{2xy - 1}} + \dfrac{{{x^2} - 3}}{{x + 2}}\\
= - x - 2 + \dfrac{{{x^2} - 3}}{{x + 2}}\\
= \dfrac{{ - {x^2} - 2x - 2x - 4 + {x^2} - 3}}{{x + 2}}\\
= \dfrac{{ - 4x - 7}}{{x + 2}}
\end{array}\)
