hãy giúp tôi hãy giúp tôi hãy giúp tôi hãy giúp tôi hãy giúp tôi

Đáp án:

$\begin{array}{l}
Dkxd:x \ne 3;x \ne  - 3;x \ne 0;x \ne \dfrac{1}{2}\\
a)P = \left( {\dfrac{6}{{{x^2} - 9}} - \dfrac{5}{{3 - x}} + \dfrac{1}{{x + 3}}} \right):\dfrac{{2x - 1}}{{{x^2} - 3x}}\\
 = \left( {\dfrac{6}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{5}{{x - 3}} + \dfrac{1}{{x + 3}}} \right).\dfrac{{x\left( {x - 3} \right)}}{{2x - 1}}\\
 = \dfrac{{6 + 5.\left( {x + 3} \right) + x - 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}.\dfrac{{x\left( {x - 3} \right)}}{{2x - 1}}\\
 = \dfrac{{6 + 5x + 15 + x - 3}}{{x + 3}}.\dfrac{x}{{2x - 1}}\\
 = \dfrac{{6x + 18}}{{x + 3}}.\dfrac{x}{{2x - 1}}\\
 = \dfrac{{6x}}{{2x - 1}}\\
b)x = \dfrac{1}{3}\left( {tm} \right)\\
 \Leftrightarrow P = \dfrac{{6.\dfrac{1}{3}}}{{2.\dfrac{1}{3} - 1}} = \dfrac{2}{{\dfrac{2}{3} - 1}} = \dfrac{2}{{ - \dfrac{1}{3}}} =  - 6\\
c)P = \dfrac{{6x}}{{2x - 1}}\\
 = \dfrac{{6x - 3 + 3}}{{2x - 1}}\\
 = \dfrac{{3.\left( {2x - 1} \right) + 3}}{{2x - 1}}\\
 = 3 + \dfrac{3}{{2x - 1}}\\
P \in Z\\
 \Leftrightarrow \dfrac{3}{{2x - 1}} \in Z\\
 \Leftrightarrow \left( {2x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Leftrightarrow 2x \in \left\{ { - 2;0;2;4} \right\}\\
 \Leftrightarrow x \in \left\{ { - 1;0;1;2} \right\}\\
Do:x \ne 0\\
 \Leftrightarrow x \in \left\{ { - 1;1;2} \right\}
\end{array}$