Đáp án:
2) \(\dfrac{{3x}}{{x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = - \dfrac{1}{3}\\
\to Q = \dfrac{{2.\left( { - \dfrac{1}{3}} \right)}}{{ - \dfrac{1}{3} + 3}} = - \dfrac{1}{4}\\
2)P = \dfrac{{x + 1}}{{x - 3}} + \dfrac{{11x - 3}}{{{x^2} - 9}} = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right) + 11x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} + 4x + 3 + 11x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} + 15x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
M = Q + P = \dfrac{{{x^2} + 15x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{2x}}{{x + 3}}\\
= \dfrac{{{x^2} + 15x + 2x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{{x^2} + 15x + 2{x^2} - 6x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3{x^2} + 9x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{3x}}{{x - 3}}\\
3)M = \dfrac{{3x}}{{x - 3}} = \dfrac{{3\left( {x - 3} \right) + 9}}{{x - 3}} = 3 + \dfrac{9}{{x - 3}}\\
M \in Z\\
\Leftrightarrow \dfrac{9}{{x - 3}} \in Z\\
\Leftrightarrow x - 3 \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
x - 3 = 9\\
x - 3 = - 9\\
x - 3 = 3\\
x - 3 = - 3\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x = - 6\\
x = 6\\
x = 0\\
x = 4\\
x = 2
\end{array} \right.
\end{array}\)
