Đáp án:
d) \(\left[ \begin{array}{l}
x = 8\\
x = - 14\\
x = - 2\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{2x - 5}}{{x + 3}}\\
B = \dfrac{{3x}}{{3 - x}} + \dfrac{{{x^2}}}{{{x^2} - 9}}\\
= \dfrac{{{x^2} - 3x\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{ - 2{x^2} - 9x}}{{{x^2} - 9}}\\
M = A + B = \dfrac{{2x - 5}}{{x + 3}} + \dfrac{{ - 2{x^2} - 9x}}{{{x^2} - 9}}\\
= \dfrac{{\left( {2x - 5} \right)\left( {x - 3} \right) - 2{x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{2{x^2} - 11x + 15 - 2{x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{ - 11x + 6}}{{{x^2} - 9}}\\
b)Thay:x = 1\\
\to A = \dfrac{{2.1 - 5}}{{1 + 3}} = - \dfrac{3}{4}\\
c)A = - \dfrac{2}{3}\\
\to \dfrac{{2x - 5}}{{x + 3}} = - \dfrac{2}{3}\\
\to - 2x - 6 = 6x - 15\\
\to 8x = 9\\
\to x = \dfrac{9}{8}\\
d)A = \dfrac{{2x - 5}}{{x + 3}} = \dfrac{{2\left( {x + 3} \right) - 11}}{{x + 3}}\\
= 2 - \dfrac{{11}}{{x + 3}}\\
A \in Z\\
\to \dfrac{{11}}{{x + 3}} \in Z\\
\to x + 3 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
x + 3 = 11\\
x + 3 = - 11\\
x + 3 = 1\\
x + 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 8\\
x = - 14\\
x = - 2\\
x = - 4
\end{array} \right.
\end{array}\)
