Đáp án:
\(\begin{array}{l}
a)A = \dfrac{{3x}}{{x - 3}}\\
b)x = - \dfrac{{15}}{7}\\
c)x > - 1;x \ne 3\\
d)\left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3; - 1;3} \right\}\\
A = \dfrac{{2x}}{{x + 3}} + \dfrac{{x + 1}}{{x - 3}} + \dfrac{{3 - 11x}}{{9 - {x^2}}}\\
= \dfrac{{2x\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x + 3} \right) - 3 + 11x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{2{x^2} - 6x + {x^2} + 4x + 3 - 3 + 11x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3{x^2} + 9x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3x}}{{x - 3}}\\
b)B = \dfrac{9}{2}\\
\to \dfrac{{x - 3}}{{x + 1}} = \dfrac{9}{2}\\
\to 2x - 6 = 9x + 9\\
\to 7x = - 15\\
\to x = - \dfrac{{15}}{7}\\
c)B < 1\\
\to \dfrac{{x - 3}}{{x + 1}} < 1\\
\to \dfrac{{x - 3 - x - 1}}{{x + 1}} < 0\\
\to \dfrac{{ - 4}}{{x + 1}} < 0\\
\to x + 1 > 0\\
\to x > - 1;x \ne 3\\
d)P = A.B = \dfrac{{3x}}{{x - 3}}.\dfrac{{x - 3}}{{x + 1}}\\
= \dfrac{{3x}}{{x + 1}} = \dfrac{{3\left( {x + 1} \right) - 3}}{{x + 1}}\\
= 3 - \dfrac{3}{{x + 1}}\\
P \in Z \to \dfrac{3}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = - 3\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
